Evaluating 0/0 form

The \(0/0\) form is one of the most significant forms of limits and is often referred to as the "mother" of all indeterminate forms. This is because many other indeterminate forms can eventually be reduced to this form, and the methods we learn here to evaluate \(0/0\) limits are universally applicable to those other forms as well.

Definition of the \(0/0\) Form

Consider the limit \(\lim_{x \to a} \frac{f(x)}{g(x)}\), where both \(f(x)\) and \(g(x)\) approach \(0\) as \(x \to a\):

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}. \]

This is an indeterminate form because its value cannot be concluded directly and can vary depending on the behavior of the functions \(f(x)\) and \(g(x)\). For instance:

• The limit could lead to a finite value, such as \(2\) or \(-3\).
• The limit could diverge to \(\infty\) or \(-\infty\).
• The limit might oscillate, failing to settle on any value.

Since the outcome depends entirely on the interplay between \(f(x)\) and \(g(x)\), further analysis is required to resolve this indeterminate form.

Factorization

The first method for evaluating \(0/0\) limits is factorization. This approach is especially useful when both \(f(x)\) and \(g(x)\) are polynomials. If \(f(x)\) and \(g(x)\) both vanish at \(x = a\), then \(x - a\) must be a common factor. By factoring \(f(x)\) and \(g(x)\), we can cancel the terms causing the \(0/0\) form, effectively removing the part of the expression responsible for the indeterminacy. The remaining expression is often determinate, allowing the limit to be evaluated directly.

Let us understand through examples:

Example:

Evaluate \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2}\).

Step 1: Check the form.

As \(x \to 2\):

\[ f(x) = x^2 - 4 \to 0, \quad g(x) = x - 2 \to 0. \]

Thus, the limit is of the form \(\frac{0}{0}\), requiring further analysis.

Step 2: Factorize.

Factor \(f(x) = x^2 - 4\) as \((x - 2)(x + 2)\):

\[ \frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}. \]

Step 3: Cancel common terms.

Cancel the factor \((x - 2)\), which is causing the \(0/0\) form:

\[ \frac{(x - 2)(x + 2)}{x - 2} = x + 2 \quad \text{for } x \neq 2. \]

Step 4: Evaluate the simplified expression.

Now the limit becomes:

\[ \lim_{x \to 2} (x + 2) = 2 + 2 = 4. \]

Thus, The value of the limit is \(4\). By removing the factor causing the \(0/0\) form, we transformed the problem into a determinate one and evaluated the limit.

The \(0/0\) form arises because of the simultaneous vanishing of \(f(x)\) and \(g(x)\). This behavior is often due to specific factors in \(f(x)\) and \(g(x)\) that become zero at the same point. Factorization provides a systematic way to cancel such terms and simplify the expression. Once the indeterminate part is removed, the limit typically becomes determinate and can be evaluated directly.

This method is particularly effective for polynomial functions, and it sets the foundation for handling more complex limits. In the next sections, we will explore additional methods, such as rationalization and L’Hôpital’s Rule, to further extend our ability to resolve \(0/0\) forms and other indeterminate forms.

Rationalization

When the \(0/0\) form involves radicals, particularly expressions of the form \(p(x)^{1/n} - q(x)^{1/n}\) that are approaching \(0\)contributing to \(0/0\) form, rationalization becomes a useful method. By multiplying both the numerator and denominator by the conjugate of the numerator, we can eliminate the radicals, simplifying the expression. This allows further evaluation through factorization or other methods.

Example

Evaluate \(\lim_{x \to 1} \frac{\sqrt{1 + x + x^2} - \sqrt{2 + x^2}}{x - 1}\).

Step 1: Check the form.

As \(x \to 1\):

• Numerator: \(\sqrt{1 + x + x^2} - \sqrt{2 + x^2} \to \sqrt{3} - \sqrt{3} = 0\),
• Denominator: \(x - 1 \to 0\).

Thus, the limit is of the form \(\frac{0}{0}\).

Step 2: Rationalize the numerator.

Multiply both numerator and denominator by the conjugate of the numerator, \(\sqrt{1 + x + x^2} + \sqrt{2 + x^2}\):

\[ \lim_{x \to 1} \frac{\sqrt{1 + x + x^2} - \sqrt{2 + x^2}}{x - 1} \cdot \frac{\sqrt{1 + x + x^2} + \sqrt{2 + x^2}}{\sqrt{1 + x + x^2} + \sqrt{2 + x^2}}. \]

This simplifies to:

\[ \lim_{x \to 1} \frac{\left(1 + x + x^2\right) - \left(2 + x^2\right)}{(x - 1)\left(\sqrt{1 + x + x^2} + \sqrt{2 + x^2}\right)}. \]

Step 3: Simplify the numerator.

The numerator becomes:

\[ \left(1 + x + x^2\right) - \left(2 + x^2\right) = 1 + x + x^2 - 2 - x^2 = x - 1. \]

Substitute back into the limit:

\[ \lim_{x \to 1} \frac{x - 1}{(x - 1)\left(\sqrt{1 + x + x^2} + \sqrt{2 + x^2}\right)}. \]

Step 4: Cancel the common factor.

Cancel \(x - 1\) in the numerator and denominator (valid for \(x \neq 1\)):

\[ \lim_{x \to 1} \frac{1}{\sqrt{1 + x + x^2} + \sqrt{2 + x^2}}. \]

Step 5: Evaluate the simplified expression.

As \(x \to 1\):

\[ \sqrt{1 + x + x^2} \to \sqrt{1 + 1 + 1^2} = \sqrt{3}, \quad \sqrt{2 + x^2} \to \sqrt{2 + 1^2} = \sqrt{3}. \]

Thus:

\[ \lim_{x \to 1} \frac{1}{\sqrt{1 + x + x^2} + \sqrt{2 + x^2}} = \frac{1}{\sqrt{3} + \sqrt{3}} = \frac{1}{2\sqrt{3}} = \frac{\sqrt{3}}{6}. \]

Rationalizing Factors for \(x^{1/n} - y^{1/n}\) and \(x^{1/n} + y^{1/n}\)

When working with radicals of the form \(x^{1/n} - y^{1/n}\) or \(x^{1/n} + y^{1/n}\), rationalizing factors can be used to eliminate the radicals and simplify expressions. Below is a detailed explanation of the rationalizing factors for these cases.

Case 1: \(x^{1/n} - y^{1/n}\) for \(n = 2, 3, 4, \dots\)

The rationalizing factor of \(x^{1/n} - y^{1/n}\) is given by:

\[ a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}, \]

where \(a = x^{1/n}\) and \(b = y^{1/n}\).

When multiplied, this satisfies:

\[ (x^{1/n} - y^{1/n}) \cdot (a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}) = x - y. \]

Example (for \(n = 3\)):

The rationalizing factor for \(x^{1/3} - y^{1/3}\) is:

\[ x^{2/3} + x^{1/3}y^{1/3} + y^{2/3}. \]

Thus:

\[ (x^{1/3} - y^{1/3})(x^{2/3} + x^{1/3}y^{1/3} + y^{2/3}) = x - y. \]

---

Case 2: \(x^{1/n} + y^{1/n}\) for odd integers \(n = 3, 5, 7, \dots\)

The rationalizing factor of \(x^{1/n} + y^{1/n}\) is given by:

\[ a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \cdots - ab^{n-2} + b^{n-1}, \]

where \(a = x^{1/n}\) and \(b = y^{1/n}\).

When multiplied, this satisfies:

\[ (x^{1/n} + y^{1/n}) \cdot (a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \cdots - ab^{n-2} + b^{n-1}) = x + y. \]

Example (for \(n = 3\)):

The rationalizing factor for \(x^{1/3} + y^{1/3}\) is:

\[ x^{2/3} - x^{1/3}y^{1/3} + y^{2/3}. \]

Thus:

\[ (x^{1/3} + y^{1/3})(x^{2/3} - x^{1/3}y^{1/3} + y^{2/3}) = x + y. \]

For \(n\) we cannot rationalize \(x^{1/n} + y^{1/n}\)

Example

Evaluate \(\lim_{x \to 0} \frac{(1+x)^{1/5} - 1}{x}\).

Solution:

First, observe that as \(x \to 0\):

• Numerator: \((1+x)^{1/5} - 1 \to 1 - 1 = 0\),
• Denominator: \(x \to 0\).

Thus, the limit is of the form \(\frac{0}{0}\), which is an indeterminate form. To resolve this, we use rationalization to simply the numerator first.

Step 1: Rationalizing the Numerator

The rationalizing factor for \((1+x)^{1/5} - 1\) is:

\[ (1+x)^{4/5} + (1+x)^{3/5} + (1+x)^{2/5} + (1+x)^{1/5} + 1. \]

Multiply numerator and denominator by this factor:

\[ \lim_{x \to 0} \frac{(1+x)^{1/5} - 1}{x} = \lim_{x \to 0} \frac{\left((1+x)^{1/5} - 1\right) \left((1+x)^{4/5} + (1+x)^{3/5} + (1+x)^{2/5} + (1+x)^{1/5} + 1\right)}{x \left((1+x)^{4/5} + (1+x)^{3/5} + (1+x)^{2/5} + (1+x)^{1/5} + 1\right)}. \]

The numerator simplifies using the identity:

\[ \left((1+x)^{1/5} - 1\right)\left((1+x)^{4/5} + (1+x)^{3/5} + (1+x)^{2/5} + (1+x)^{1/5} + 1\right) = (1+x) - 1 = x. \]

Thus, the expression becomes:

\[ \lim_{x \to 0} \frac{x}{x \left((1+x)^{4/5} + (1+x)^{3/5} + (1+x)^{2/5} + (1+x)^{1/5} + 1\right)}. \]

Step 2: Simplify and Evaluate

Cancel \(x\) from numerator and denominator as \(x\) is approaching \(0\) and hence \(\ne 0\)

\[ \lim_{x \to 0} \frac{1}{(1+x)^{4/5} + (1+x)^{3/5} + (1+x)^{2/5} + (1+x)^{1/5} + 1}. \]

As \(x \to 0\), \((1+x)^{k/5} \to 1\) for \(k = 1, 2, 3, 4\). Substituting these limits:

\[ \lim_{x \to 0} \frac{1}{1 + 1 + 1 + 1 + 1} = \frac{1}{5}. \]

therefore,

The value of the limit is:

\[ \lim_{x \to 0} \frac{(1+x)^{1/5} - 1}{x} = \frac{1}{5}. \]

Example

Evaluate \(\lim_{x \to 1} \frac{x^2 - 1}{|x - 1|}\).

Solution:

The absolute value function \(|x - 1|\) is defined piecewise as follows:

\[ |x - 1| = \begin{cases} x - 1 & \text{if } x \geq 1, \\ -(x - 1) & \text{if } x < 1. \end{cases} \]

Thus, the limit depends on whether \(x\) approaches \(1\) from the left (\(x \to 1^-\)) or the right (\(x \to 1^+\)).

Left-Hand Limit (\(x \to 1^-\))

When \(x \to 1^-\), \(x < 1\) implies \(x - 1 < 0\), so \(|x - 1| = -(x - 1)\). Substituting this into the expression, we have:

\[ \lim_{x \to 1^-} \frac{x^2 - 1}{|x - 1|} = \lim_{x \to 1^-} \frac{x^2 - 1}{-(x - 1)}. \]

Factoring \(x^2 - 1\) as \((x - 1)(x + 1)\), this becomes:

\[ \lim_{x \to 1^-} \frac{(x - 1)(x + 1)}{-(x - 1)} = \lim_{x \to 1^-} -(x + 1). \]

As \(x \to 1^-\), \(x + 1 \to 2\), so:

\[ \lim_{x \to 1^-} \frac{x^2 - 1}{|x - 1|} = -2. \]

Right-Hand Limit (\(x \to 1^+\))

When \(x \to 1^+\), \(x > 1\) implies \(x - 1 > 0\), so \(|x - 1| = x - 1\). Substituting this into the expression, we have:

\[ \lim_{x \to 1^+} \frac{x^2 - 1}{|x - 1|} = \lim_{x \to 1^+} \frac{x^2 - 1}{x - 1}. \]

Factoring \(x^2 - 1\) as \((x - 1)(x + 1)\), this becomes:

\[ \lim_{x \to 1^+} \frac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1^+} x + 1. \]

As \(x \to 1^+\), \(x + 1 \to 2\), so:

\[ \lim_{x \to 1^+} \frac{x^2 - 1}{|x - 1|} = 2. \]

Hence,

The left-hand limit and right-hand limit are not equal:

\[ \lim_{x \to 1^-} \frac{x^2 - 1}{|x - 1|} = -2, \quad \lim_{x \to 1^+} \frac{x^2 - 1}{|x - 1|} = 2. \]

Therefore, the two-sided limit does not exist:

\[ \lim_{x \to 1} \frac{x^2 - 1}{|x - 1|} \text{ does not exist.} \]

Example

Evaluate \(a + b\) if \(\lim_{x \to 1} \frac{x^2 - ax + b}{x - 1} = 10\).

Solution:

The given limit involves a rational function. As \(x \to 1\), the denominator \(x - 1 \to 0\), and for the limit to exist as a finite value, the numerator must also approach 0. Let the numerator be \(x^2 - ax + b\).

As \(x \to 1\), the numerator approaches:

\[ 1 - a + b. \]

If \(1 - a + b \neq 0\), then the limit would tend to \(\pm \infty\) because of division by 0. Since the given limit is finite (\(10\)), it is necessary that \(1 - a + b = 0\), giving:

\[ b = a - 1. \]

Substituting \(b = a - 1\) into the numerator, the limit becomes:

\[ \lim_{x \to 1} \frac{x^2 - ax + a - 1}{x - 1}. \]

Factorize the numerator. Group terms to extract \((x - 1)\) as a factor:

\[ x^2 - ax + a - 1 = (x - 1)(x + 1 - a). \]

Thus, the expression becomes:

\[ \lim_{x \to 1} \frac{(x - 1)(x + 1 - a)}{x - 1}. \]

Cancel the common factor of \(x - 1\) (valid since \( x \to 1 \implies\) \(x \neq 1\)):

\[ \lim_{x \to 1} (x + 1 - a). \]

Substitute \(x = 1\) into the remaining expression:

\[ 1 + 1 - a = 10. \]

Simplify:

\[ 2 - a = 10 \implies a = -8. \]

Now, using \(b = a - 1\), substitute \(a = -8\):

\[ b = -8 - 1 = -9. \]

Thus:

\[ a + b = -8 + (-9) = -17. \]

Expansions

A power series is a representation of a function as an infinite sum of terms involving powers of a variable. It takes the form:

\[ f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots + a_nx^n + \cdots = \sum_{n=0}^\infty a_nx^n, \]

where \(a_0, a_1, a_2, \dots\) are constants, called the coefficients of the series. The key idea of a power series is that it approximates functions by expressing them as "polynomials" of infinite degree.

A Maclaurin series is a special case of the power series where the coefficients \(a_n\) are determined by the derivatives of the function \(f(x)\) at \(x = 0\). It is given by:

\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n. \]

This representation is especially useful for analyzing and approximating functions around \(x = 0\).

The power of Maclaurin series lies in its ability to simplify calculations. For instance, if you encounter a challenging limit or need a numerical approximation for a transcendental function, substituting its Maclaurin expansion often leads to a simpler solution. Below are the Maclaurin expansions for several important functions, along with their domains of validity.

Examples of Maclaurin Series:

1. Exponential Function \(e^x\):

   The series for \(e^x\) is:

   \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, \quad \text{valid for all } x. \]

   For any value of \(x\), whether positive, negative, or zero, if we calculate the terms on the right-hand side and add them up, the result will always equal \(e^x\).

   For instance, if we let \(x = 1\), substituting \(x\) into the series gives:

   \[ e^1 = 1 + 1 + \frac{1^2}{2!} + \frac{1^3}{3!} + \frac{1^4}{4!} + \cdots. \]

   Simplifying the terms, we have:

   \[ e = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \cdots. \]

   As we add more and more terms, the result gets closer and closer to the value of \(e\), which is approximately \(2.718\). But the remarkable fact is that, if we could sum infinitely many terms, the two sides would match exactly.

   Now consider \(x = -1\). Substituting \(x = -1\) into the series gives:

   \[ e^{-1} = 1 - 1 + \frac{(-1)^2}{2!} - \frac{(-1)^3}{3!} + \frac{(-1)^4}{4!} - \cdots. \]

   This simplifies to:

   \[ e^{-1} = 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \cdots. \]

   When all terms are summed, the result is exactly \(\frac{1}{e}\), which is approximately \(0.367\).

   We can also replace \(x\) with \(-x\) in the series to get:

   \[ e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \cdots. \]

   This gives the exact value of \(e^{-x}\) for any value of \(x\). For example, if \(x = 2\), the right-hand side computes \(e^{-2}\).

   This power series for \(e^x\) not only works for all values of \(x\), but it also simplifies calculation of limits:

   \[ \lim_{x \to 0} \frac{e^x - 1}{x}, \]

   we can substitute the series \(e^x = 1 + x + \frac{x^2}{2!} + \cdots\). The numerator becomes:

   \[ e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. \]

   Dividing by \(x\), we get:

   \[ \frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots. \]

   As \(x \to 0\), all higher powers of \(x\) vanish, leaving:

   \[ \frac{e^x - 1}{x} \to 1. \]

   This result shows how the series makes the calculation almost effortless.

   Thus, the exponential series is not only valid for any value of \(x\), but it also serves as a practical tool for approximating values, solving limits, and understanding deeper mathematical properties.

   proof for the Maclaurin's series for \(e^x\)

   To derive the Maclaurin series for \(e^x\), we start with the definition of the Maclaurin series:

   \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots. \]

   For \(f(x) = e^x\), we know that all derivatives of \(e^x\) are equal to \(e^x\). At \(x = 0\), \(f^{(n)}(0) = e^0 = 1\) for all \(n\). Substituting into the Maclaurin series formula:

   \[ e^x = 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \cdots + \frac{1}{n!}x^n + \cdots = \sum_{n=0}^\infty \frac{x^n}{n!}. \]

   This shows that the exponential function \(e^x\) can be represented as this infinite series, and the series holds true for all values of \(x\), as \(e^x\) is smooth and infinitely differentiable everywhere.

2. General Exponential Function \(a^x\):

   For \(a^x\), the expansion is:

   \[ a^x = e^{x\ln x} = 1 + x \ln a + \frac{x^2 (\ln a)^2}{2!} + \frac{x^3 (\ln a)^3}{3!} + \cdots, \quad \text{valid for } a > 0. \]

3. Logarithmic Function \(\ln(1 + x)\):

   The expansion for \(\ln(1 + x)\) is:

   \[ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots, \quad -1 < x \leq 1. \]

   This series only works within the given range of \(x\).

   Substituting \(x = 1\) into the series, we get:

   \[ \boxed{\ln(2) = \ln(1+1) = 1 - \frac{1^2}{2} + \frac{1^3}{3} - \frac{1^4}{4} + \cdots.} \]

   The Maclaurin series for \(\ln(1-x)\) is derived by putting \(-x\) in place of \(x\) in \(\ln(1+x)\), and it is given by:

   \[ \ln(1-x) = - \left( x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots \right) = -x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots, \]

   valid for \( -1 \leq x < 1\).

4. Sine Function \(\sin x\):

   The Maclaurin series for \(\sin x\) is:

   \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots, \quad \text{valid for all } x. \]

   A comparision of \(\sin x\) with its polynomial approximations:

   

   This graph illustrates a comparison between the function \(\sin x\) (red curve) and its polynomial approximations derived from the Maclaurin series expansion:

   \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]

   1. Linear Approximation (\(x\)): The blue line represents the first-degree approximation \(x\), which corresponds to the leading term of the series. This approximation is accurate only very close to \(x = 0\), as it neglects all higher-order terms, leading to significant divergence from \(\sin x\) as \(|x|\) increases.

   2. Cubic Approximation (\(x - \frac{x^3}{3!}\)): The green curve represents the third-degree approximation \(x - \frac{x^3}{3!}\). This captures more of the behavior of \(\sin x\) by accounting for the cubic term. As a result, it better approximates \(\sin x\) over a larger interval, particularly for \(|x| \leq \frac{\pi}{2}\). However, it still deviates significantly for \(|x| > \frac{\pi}{2}\), as higher-order terms are ignored.

   3. Quintic Approximation (\(x - \frac{x^3}{3!} + \frac{x^5}{5!}\)): The black curve represents the fifth-degree approximation, which incorporates the next term in the Maclaurin series. This further improves the accuracy of the approximation over the interval \([- \pi, \pi]\), closely matching \(\sin x\) within this range.

   The higher the degree of the polynomial approximation, the better the alignment with \(\sin x\) over a wider interval. However, outside the plotted interval, even the quintic approximation will eventually diverge, as the series converges to \(\sin x\) only when all terms are included. This graph rigorously demonstrates how the inclusion of additional terms from the Maclaurin series progressively improves the accuracy of the approximation, reflecting the power of the series to model transcendental functions.

5. Cosine Function \(\cos x\):

   For \(\cos x\), the expansion is:

   \[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots, \quad \text{valid for all } x. \]

   

6. Tangent Function \(\tan x\):

   The Maclaurin expansion for \(\tan x\) is:

   \[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \frac{17x^7}{315} + \cdots, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}. \]

7. Inverse Tangent Function \(\tan^{-1} x\):

   The expansion for \(\tan^{-1} x\) is:

   \[ \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots, \quad |x| \leq 1. \]

8. Inverse Sine Function \(\sin^{-1} x\):

   For \(\sin^{-1} x\), the series is:

   \[ \sin^{-1} x = x + \frac{x^3}{3 \cdot 2^2} + \frac{3x^5}{5 \cdot 2^4} + \frac{5x^7}{7 \cdot 2^6} + \cdots, \quad |x| \leq 1. \]

9. Secant Function \(\sec x\):

   The series for \(\sec x\) is:

   \[ \sec x = 1 + \frac{x^2}{2!} + \frac{5x^4}{4!} + \frac{61x^6}{6!} + \cdots, \quad |x| < \frac{\pi}{2}. \]

10. Binomial Expansion \((1 + x)^n\):

    The series for \((1 + x)^n\) is:

    \[ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots, \quad |x| < 1, \; n \in \mathbb{Q}. \]

11. Generalized Root Function \((1 + x)^{1/x}\):

    For small \(x\), the series is:

    \[ (1 + x)^{1/x} = e \left(1 - \frac{x}{2} + \frac{11x^2}{24} + \cdots \right). \]

Maclaurin series are incredibly practical tools. For example, when evaluating a limit like:

\[ \lim_{x \to 0} \frac{\sin x}{x}, \]

substituting the expansion \(\sin x = x - \frac{x^3}{3!} + \cdots\) makes the solution immediate:

\[ \lim_{x \to 0} \frac{x - \frac{x^3}{3!} + \cdots}{x} = \lim_{x \to 0} \left(1 - \frac{x^2}{6} + \cdots\right) = 1. \]

Example

Evaluate \(\lim_{x \to 0} \frac{e^x - 1 - x}{\sin^2 x}\).

Solution:

As \(x \to 0\), both the numerator \(e^x - 1 - x\) and the denominator \(\sin^2 x\) approach \(0\), resulting in an indeterminate form \(\frac{0}{0}\). To evaluate this limit, we substitute the Maclaurin expansions of \(e^x\) and \(\sin x\) directly into the limit expression.

The Maclaurin series expansions are:

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, \]

and

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots. \]

Substituting these expansions into the limit, we have:

\[ \lim_{x \to 0} \frac{e^x - 1 - x}{\sin^2 x} = \lim_{x \to 0} \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right) - 1 - x}{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)^2}. \]

In the numerator, the terms \(1\) and \(x\) cancel, leaving:

\[ \lim_{x \to 0} \frac{\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{\left(x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\right)^2}. \]

Factoring \(x^2\) from the denominator:

\[ \lim_{x \to 0} \frac{\frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x^2 \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots\right)^2}. \]

Canceling \(x^2\) from the numerator and denominator:

\[ \lim_{x \to 0} \frac{\frac{1}{2!} + \frac{x}{3!} + \cdots}{\left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots\right)^2}. \]

As \(x \to 0\), all higher-order terms vanish, so the limit reduces to:

\[ \frac{\frac{1}{2!}}{1^2} = \frac{1}{2}. \]

Therefore,

\[ \lim_{x \to 0} \frac{e^x - 1 - x}{\sin^2 x} = \frac{1}{2}. \]

Example

Evaluate \(\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}\).

Solution:

As \(x \to 0\), both the numerator \(\tan x - \sin x\) and the denominator \(x^3\) approach 0, resulting in the indeterminate form \(\frac{0}{0}\). To resolve this, we use the Maclaurin expansions of \(\tan x\) and \(\sin x\) and substitute them directly into the limit.

The expansion of \(\tan x\) is:

\[ \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \cdots, \]

and the expansion of \(\sin x\) is:

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots. \]

Substituting these into the limit, we get:

\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\left(x + \frac{x^3}{3} + \cdots\right) - \left(x - \frac{x^3}{3!} + \cdots\right)}{x^3}. \]

Simplifying the numerator:

\[ \tan x - \sin x = \frac{x^3}{3} + \frac{x^3}{3!} + \cdots = x^3 \left(\frac{1}{3} + \frac{1}{6} + \cdots\right). \]

Thus, the limit becomes:

\[ \lim_{x \to 0} \frac{x^3 \left(\frac{1}{3} + \frac{1}{6} + \cdots\right)}{x^3}. \]

Canceling \(x^3\):

\[ \lim_{x \to 0} \left(\frac{1}{3} + \frac{1}{6} + \cdots\right) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}. \]

Therefore,

\[ \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \frac{1}{2}. \]

Standard Limits

Some specific limit forms appear so frequently in problems that it is essential to recognize and remember them as standard limits. These limits often simplify expressions and help resolve indeterminate forms directly. By understanding these standard patterns, we can handle problems more efficiently without deriving each result from scratch.

These standard limits are not isolated facts; they frequently appear in problems involving basic operations with functions like trigonometric, exponential, or logarithmic expressions. Knowing them allows us to quickly evaluate or transform problems into simpler forms. We will see how these limits arise naturally in specific cases and learn how to apply them effectively in various problems.

Standard Limit I:

\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}, \, n \in \mathbb{Q}. \]

Proof:

As \(x \to a \implies x - a \to 0\).
Substitute \(x - a = h\), so \(h \to 0\).

\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = \lim_{h \to 0} \frac{(a + h)^n - a^n}{h}. \]

\[ = \lim_{h \to 0} \frac{a^n \left(1 + \frac{h}{a}\right)^n - a^n}{h}. \]

Using the infinite expansion of \((1 + x)^n\), valid for \(n \in \mathbb{Q}\):

\[ (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots, \]

substitute \(x = \frac{h}{a}\), giving:

\[ \lim_{h \to 0} \frac{a^n \left(1 + \frac{n h}{a} + \frac{n(n-1)}{2!} \frac{h^2}{a^2} + \cdots\right) - a^n}{h}. \]

Simplify:

\[ \lim_{h \to 0} \frac{a^n \left(\frac{n h}{a} + \frac{n(n-1)}{2!} \frac{h^2}{a^2} + \cdots\right)}{h}. \]

Factor \(h\) from the numerator:

\[ \lim_{h \to 0} a^n \left(\frac{n}{a} + \frac{n(n-1)}{2!} \frac{h}{a^2} + \cdots\right). \]

As \(h \to 0\), all terms involving \(h\) vanish, leaving:

\[ a^n \cdot \frac{n}{a} = n a^{n-1}. \]

Hence,

\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}. \]

\(\blacksquare\)

Example

Evaluate

\[ \lim_{x \to 8} \frac{x^{1/3} - 2}{x - 8}. \]

Solution:

The given limit is of the indeterminate form \(\frac{0}{0}\). Recognizing the standard form,

\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}, \quad n \in \mathbb{Q}, \]

with \( n = \frac{1}{3} \) and \( a = 8 \), applying the result gives

\[ \lim_{x \to 8} \frac{x^{1/3} - 2}{x - 8} = \frac{1}{3} \cdot 8^{-2/3}. \]

Since \( 8^{-2/3} = \frac{1}{4} \),

\[ \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}. \]

Thus,

\[ \lim_{x \to 8} \frac{x^{1/3} - 2}{x - 8} = \frac{1}{12}. \]

Example

Evaluate

\[ \lim_{x \to -1} \frac{1 + x^{1/7}}{1 + x^{1/11}}. \]

Solution:

The limit is in the indeterminate form \(\frac{0}{0}\), so dividing both numerator and denominator by \(x + 1\):

\[ \lim_{x \to -1} \frac{\frac{1 + x^{1/7}}{x + 1}}{\frac{1 + x^{1/11}}{x + 1}} = \lim_{x \to -1} \frac{\frac{x^{1/7} - (-1)^{1/7}}{x - (-1)}}{\frac{x^{1/11} - (-1)^{1/11}}{x - (-1)}}. \]

Applying the standard limit

\[ \lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}, \]

with \( a = -1 \), \( n = \frac{1}{7} \) in the numerator, and \( n = \frac{1}{11} \) in the denominator, we get:

\[ \frac{\frac{1}{7} (-1)^{-6/7}}{\frac{1}{11} (-1)^{-10/11}} = \frac{1}{7} (-1)^{-6/7} \times \frac{11}{1} (-1)^{10/11}. \]

\[ = \frac{11}{7}. \]

Example

Prove that

\[ \lim_{x \to 0} \frac{(1 + x)^n - 1}{x} = n. \]

Proof:

Substituting \( y = 1 + x \), we get \( x = y - 1 \), and as \( x \to 0 \), it follows that \( y \to 1 \). Rewriting the given limit in terms of \( y \):

\[ \lim_{x \to 0} \frac{(1 + x)^n - 1}{x} = \lim_{y \to 1} \frac{y^n - 1}{y - 1}. \]

Recognizing the standard limit,

\[ \lim_{y \to a} \frac{y^n - a^n}{y - a} = n a^{n-1}, \]

with \( a = 1 \), applying this result gives

\[ \lim_{y \to 1} \frac{y^n - 1}{y - 1} = n \cdot 1^{n-1} = n. \]

Thus,

\[ \lim_{x \to 0} \frac{(1 + x)^n - 1}{x} = n. \]

Standard Limit II:

\[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1. \]

Proof:

Using the Maclaurin series expansion of \( e^x \):

\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. \]

Substituting into the given limit:

\[ \lim_{x \to 0} \frac{e^x - 1}{x} = \lim_{x \to 0} \frac{\left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\right) - 1}{x}. \]

Canceling 1:

\[ \lim_{x \to 0} \frac{x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots}{x}. \]

Factoring \( x \):

\[ \lim_{x \to 0} \left( 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots \right). \]

As \( x \to 0 \), all higher-order terms vanish, leaving:

\[ 1. \]

Thus,

\[ \lim_{x \to 0} \frac{e^x - 1}{x} = 1. \]

\(\blacksquare\)

Example

Evaluate

\[ \lim_{x \to 1} \frac{e^x - e}{x - 1}. \]

Solution:
Clearly, substituting \( x = 1 \) gives \( \frac{e^1 - e}{1 - 1} = \frac{0}{0} \), an indeterminate form. To resolve this, factor \( e \) from the numerator:

\[ \lim_{x \to 1} \frac{e^x - e}{x - 1} = \lim_{x \to 1} \frac{e (e^{x-1} - 1)}{x - 1}. \]

Introduce \( t = x - 1 \), so as \( x \to 1 \), we have \( t \to 0 \). Rewriting in terms of \( t \):

\[ \lim_{t \to 0} \frac{e (e^t - 1)}{t}. \]

Using the standard limit result:

\[ \lim_{t \to 0} \frac{e^t - 1}{t} = 1, \]

it follows that:

\[ \lim_{t \to 0} \frac{e (e^t - 1)}{t} = e \]

Thus,

\[ \lim_{x \to 1} \frac{e^x - e}{x - 1} = e. \]

General form

For any function \( f(x) \) such that \( \lim_{x \to a} f(x) = 0 \), the following limit holds:

\[ \lim_{x \to a} \frac{e^{f(x)} - 1}{f(x)} = 1. \]

This result follows from the standard limit

\[ \lim_{t \to 0} \frac{e^t - 1}{t} = 1, \]

which remains valid when \( f(x) \to 0 \) as \( x \to a \), ensuring that the function behaves analogously in the limit process.

For example, consider

\[ \lim_{x \to \frac{\pi}{2}} \frac{e^{\cos x} - 1}{\cos x}. \]

Since \(\cos x \to 0\) as \(x \to \frac{\pi}{2}\), we can apply the result

\[ \lim_{x \to a} \frac{e^{f(x)} - 1}{f(x)} = 1 \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

Here, \( f(x) = \cos x \), and since \( \cos x \to 0 \) as \( x \to \frac{\pi}{2} \), it follows that

\[ \lim_{x \to \frac{\pi}{2}} \frac{e^{\cos x} - 1}{\cos x} = 1. \]

Standard Limit III:

For any \( a > 0 \),

\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a. \]

To prove this, use the identity

\[ a^x = e^{x \ln a}. \]

Rewriting the given limit:

\[ \lim_{x \to 0} \frac{e^{x \ln a} - 1}{x}. \]

Multiplying and dividing by \( \ln a \):

\[ \lim_{x \to 0} \ln a \cdot \frac{e^{x \ln a} - 1}{x \ln a}. \]

Applying the standard result

\[ \lim_{t \to 0} \frac{e^t - 1}{t} = 1, \]

with \( t = x \ln a \), it follows that

\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a. \]

Example

For example, consider the limit

\[ \lim_{x \to 0} \frac{3^x - 2^x}{x}. \]

To evaluate this, rewrite the numerator by factoring out \( 2^x \):

\[ 3^x - 2^x = 2^x \left( \left(\frac{3}{2}\right)^x - 1 \right), \]

which transforms the limit into

\[ \lim_{x \to 0} \frac{2^x \left( \left(\frac{3}{2}\right)^x - 1 \right)}{x}. \]

Since \( 2^x \) is continuous at \( x = 0 \) and satisfies \( \lim_{x \to 0} 2^x = 1 \), we can take it outside the limit:

\[ \lim_{x \to 0} 2^x \cdot \lim_{x \to 0} \frac{\left(\frac{3}{2}\right)^x - 1}{x}. \]

Applying the standard result

\[ \lim_{x \to 0} \frac{a^x - 1}{x} = \ln a, \]

with \( a = \frac{3}{2} \), we obtain

\[ \lim_{x \to 0} \frac{3^x - 2^x}{x} = 1 \cdot \ln \left(\frac{3}{2} \right). \]

Thus,

\[ \lim_{x \to 0} \frac{3^x - 2^x}{x} = \ln \left(\frac{3}{2} \right). \]

In general, for any positive numbers \( a, b > 0 \),

\[ \lim_{x \to 0} \frac{a^x - b^x}{x} = \ln \left( \frac{a}{b} \right). \]

Standard Limit IV:

A fundamental limit involving the natural logarithm function is

\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1. \]

This result can be established using the Maclaurin series expansion of \(\ln(1 + x)\), valid for \(|x| < 1\):

\[ \ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots. \]

Dividing both sides by \( x \) and taking the limit,

\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = \lim_{x \to 0} \left( 1 - \frac{x}{2} + \frac{x^2}{3} - \frac{x^3}{4} + \cdots \right). \]

Since all terms beyond the first vanish as \( x \to 0 \), the limit evaluates to \(1\).

Standard Limit V:

A more general form of the standard limit involving logarithms is

\[ \lim_{x \to 0} \frac{\log_a (1 + x)}{x} = \frac{1}{\ln a}, \quad a > 0, a \neq 1. \]

To establish this, recall the change of base formula for logarithms:

\[ \log_a (1 + x) = \frac{\ln(1 + x)}{\ln a}. \]

Substituting this into the given limit,

\[ \lim_{x \to 0} \frac{\log_a (1 + x)}{x} = \lim_{x \to 0} \frac{\frac{\ln(1 + x)}{\ln a}}{x}. \]

Rewriting,

\[ \frac{1}{\ln a} \lim_{x \to 0} \frac{\ln(1 + x)}{x}. \]

Using the known result

\[ \lim_{x \to 0} \frac{\ln(1 + x)}{x} = 1, \]

it follows that

\[ \lim_{x \to 0} \frac{\log_a (1 + x)}{x} = \frac{1}{\ln a}. \]

Standard Limit VI:

A classic standard limit involving trigonometric functions is

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]

Proof:

Consider the Maclaurin series expansion for \(\sin x\), which is valid for all \(x\):

\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots. \]

Substituting this series into the limit, we obtain

\[ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots}{x}. \]

Factor \(x\) from the numerator:

\[ \lim_{x \to 0} \left( 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots \right). \]

As \(x \to 0\), all higher-order terms vanish, leaving

\[ 1. \]

Therefore,

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]

Standard Limit VII:

\[ \lim_{x \to 0} \frac{\tan x}{x} = 1. \]

The identity \(\tan x = \frac{\sin x}{\cos x}\) gives

\[ \frac{\tan x}{x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x}. \]

Taking limits,

\[ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \lim_{x \to 0} \frac{1}{\cos x}. \]

Since \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\lim_{x \to 0} \frac{1}{\cos x} = 1\), the result follows:

\[ \lim_{x \to 0} \frac{\tan x}{x} = 1. \]

Standard Limit VIII:

\[ \lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1. \]

Proof:

Let \( x = \sin \theta \), so as \( x \to 0 \), it follows that \( \theta \to 0 \). Rewriting the given limit in terms of \( \theta \):

\[ \lim_{x \to 0} \frac{\sin^{-1} x}{x} = \lim_{\theta \to 0} \frac{\sin^{-1} (\sin \theta)}{\sin \theta} = \lim_{\theta \to 0} \frac{\theta}{\sin \theta}. \]

Using the standard limit

\[ \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1, \]

which implies

\[ \lim_{\theta \to 0} \frac{\theta}{\sin \theta} = 1, \]

it follows that

\[ \lim_{x \to 0} \frac{\sin^{-1} x}{x} = 1. \]

\(\blacksquare\)

Standard Limit IX

\[ \lim_{x \to 0} \frac{\tan^{-1} x}{x} = 1. \]

Using Standard Limits

These standard limits are enough to handle most problems involving limits. Whenever we see an expression that matches one of these standard forms, we can apply the result directly instead of expanding or using other complicated methods. In many cases, the given expression may not immediately look like a standard limit, but with some algebraic manipulation—such as factoring, rewriting terms, or making substitutions—we can bring it into a familiar form. Sometimes, multiple standard limits may be used together, either by multiplying, dividing, or applying them separately to different parts of the expression. Recognizing these patterns makes solving limits much faster and more straightforward.

Now, we rewrite the standard limits in a more general form using a function \( f(x) \) such that \( \lim_{x \to a} f(x) = 0 \). These general forms help recognize standard patterns in a variety of problems.

1. Exponential Function:

   \[ \lim_{x \to a} \frac{e^{f(x)} - 1}{f(x)} = 1, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

2. Logarithmic Function:

   \[ \lim_{x \to a} \frac{\ln(1 + f(x))}{f(x)} = 1, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

3. General Exponential Base:

   \[ \lim_{x \to a} \frac{a^{f(x)} - 1}{f(x)} = \ln a, \quad a > 0, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

4. Trigonometric Limits:

   \[ \lim_{x \to a} \frac{\sin f(x)}{f(x)} = 1, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]
   \[ \lim_{x \to a} \frac{\tan f(x)}{f(x)} = 1, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

5. Inverse Trigonometric Functions:

   \[ \lim_{x \to a} \frac{\sin^{-1} f(x)}{f(x)} = 1, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]
   \[ \lim_{x \to a} \frac{\tan^{-1} f(x)}{f(x)} = 1, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

6. Power Functions:

   \[ \lim_{x \to a} \frac{(1 + f(x))^n - 1}{f(x)} = n, \quad \text{if} \quad \lim_{x \to a} f(x) = 0. \]

These general forms allow us to apply standard limits more flexibly in different problems where the argument is not simply \( x \) but any function \( f(x) \) that approaches 0. By recognizing these structures within complex expressions, we can simplify limits efficiently without needing to expand terms every time.

Example

Evaluate

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2}. \]

Solution:

Using the identity \( 1 - \cos x = 2 \sin^2 (x/2) \), the given limit transforms into

\[ \lim_{x \to 0} \frac{2 \sin^2 (x/2)}{x^2}. \]

Multiplying and dividing by 4:

\[ \lim_{x \to 0} \frac{2}{4} \cdot \frac{4 \sin^2 (x/2)}{x^2}. \]

Rewriting:

\[ \frac{1}{2} \lim_{x \to 0} \left( \frac{\sin (x/2)}{x/2} \right)^2. \]

Since \( \lim_{t \to 0} \frac{\sin t}{t} = 1 \), applying this with \( t = x/2 \) gives

\[ \left( \frac{\sin (x/2)}{x/2} \right)^2 \to 1. \]

Thus,

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}. \]

You may remember this limit as it is very frequent in expressions.

Example

Evaluate

\[ \lim_{x \to 0} \frac{\sin (m x)}{\sin (n x)}. \]

Solution:

Dividing both numerator and denominator by \( x \), we write

\[ \lim_{x \to 0} \frac{\sin (m x)}{\sin (n x)} = \lim_{x \to 0} \frac{\sin (m x)}{m x} \cdot \frac{m \cancel{x}}{n \cancel{x}} \cdot \frac{n x}{\sin (n x)}. \]

Using the standard result

\[ \lim_{t \to 0} \frac{\sin t}{t} = 1, \]

it follows that

\[ \lim_{x \to 0} \frac{\sin (m x)}{m x} = 1, \quad \lim_{x \to 0} \frac{\sin (n x)}{n x} = 1. \]

Thus,

\[ \lim_{x \to 0} \frac{\sin (m x)}{\sin (n x)} = \frac{m}{n}. \]

Here is the rewritten solution with explicit cancellation of \(x\):

Example

Evaluate

\[ \lim_{x \to 0} \frac{\sqrt{1 - \cos 2x}}{x}. \]

Solution:

Substituting \( x = 0 \) gives \( \frac{0}{0} \), an indeterminate form. Using the identity

\[ 1 - \cos 2x = 2 \sin^2 x, \]

the expression transforms into

\[ \lim_{x \to 0} \frac{\sqrt{2 \sin^2 x}}{x} = \lim_{x \to 0} \frac{\sqrt{2} |\sin x|}{x}. \]

Since the modulus function appears, the left-hand limit (LHL) and right-hand limit (RHL) must be considered separately.

From the graph of \( \sin x \), it follows that:

• As \( x \to 0^+ \), \( \sin x \to 0^+ \), so \( |\sin x| = \sin x \). [because \(\sin x > 0\)]
• As \( x \to 0^- \), \( \sin x \to 0^- \), so \( |\sin x| = -\sin x \).[ because \(\sin x < 0\)]

For \( x \to 0^+ \):

\[ \lim_{x \to 0^+} \frac{\sqrt{2} \sin x}{x} = \sqrt{2} \lim_{x \to 0^+} \frac{\sin x}{x} = \sqrt{2} \cdot 1 = \sqrt{2}. \]

For \( x \to 0^- \):

\[ \lim_{x \to 0^-} \frac{\sqrt{2} (-\sin x)}{x} = \sqrt{2} \lim_{x \to 0^-} \frac{-\sin x}{x} = \sqrt{2} \cdot (-1) = -\sqrt{2}. \]

Since the left-hand and right-hand limits are not equal, the given limit does not exist.

Example

Evaluate

\[ \lim_{x \to 0} \frac{\ln(x+e) - 1}{\sin 3x}. \]

Solution:

Given,

\[ \lim_{x \to 0} \frac{\ln(x+e) - 1}{\sin 3x}. \]

Writing \(1\) as \(\ln e\),

\[ \lim_{x \to 0} \frac{\ln(x+e) - \ln e}{\sin 3x}. \]

Using the logarithm property \(\ln a - \ln b = \ln \frac{a}{b}\),

\[ \lim_{x \to 0} \frac{\ln \left( \frac{x+e}{e} \right)}{\sin 3x}. \]

Rewriting the fraction inside the logarithm,

\[ \lim_{x \to 0} \frac{\ln \left(1 + \frac{x}{e} \right)}{\sin 3x}. \]

Multiplying and dividing by \(x/e\) and \(3x\),

\[ \lim_{x \to 0} \frac{\ln \left(1 + \frac{x}{e} \right)}{\frac{x}{e}} \cdot \frac{\frac{x}{e}}{3x} \cdot \frac{3x}{\sin 3x}. \]

Evaluating each factor separately:

\[ \lim_{x \to 0} \frac{\ln \left(1 + \frac{x}{e} \right)}{\frac{x}{e}} = 1, \quad \lim_{x \to 0} \frac{3x}{\sin 3x} = 1, \quad \frac{x/e}{3x} = \frac{1}{3e}. \]

Thus,

\[ \lim_{x \to 0} \frac{\ln(x+e) - 1}{\sin 3x} = 1 \cdot \frac{1}{3e} \cdot 1 = \frac{1}{3e}. \]

Example

Evaluate

\[ \lim_{x \to 0} \frac{\ln(\cos 5x)}{x^2}. \]

Solution: Expanding \(\cos 5x\) using the identity \(\cos A = 1 - 2\sin^2 \frac{A}{2}\),

\[ \lim_{x \to 0} \frac{\ln(1 - 2\sin^2 \frac{5x}{2})}{x^2}. \]

Multiplying and dividing by \(-2\sin^2 \frac{5x}{2}\),

\[ \lim_{x \to 0} \frac{\ln(1 - 2\sin^2 \frac{5x}{2})}{-2\sin^2 \frac{5x}{2}} \cdot \frac{-2\sin^2 \frac{5x}{2}}{x^2}. \]

Rewriting the second fraction,

\[ \lim_{x \to 0} \frac{\ln(1 - 2\sin^2 \frac{5x}{2})}{-2\sin^2 \frac{5x}{2}} \cdot \left( -2 \cdot \frac{\sin^2 \frac{5x}{2}}{x^2} \right). \]

Multiplying and dividing by \( \left(\frac{5}{2} \right)^2 \),

\[ \lim_{x \to 0} \frac{\ln(1 - 2\sin^2 \frac{5x}{2})}{-2\sin^2 \frac{5x}{2}} \cdot \left( -2 \cdot \frac{25}{4} \cdot \frac{\sin^2 \frac{5x}{2}}{\left(\frac{5x}{2} \right)^2} \right). \]

Using the standard limits:

\[ \lim_{y \to 0} \frac{\ln(1+y)}{y} = 1, \quad \lim_{z \to 0} \frac{\sin^2 z}{z^2} = 1, \]

we evaluate each term:

\[ \lim_{x \to 0} \frac{\ln(1 - 2\sin^2 \frac{5x}{2})}{-2\sin^2 \frac{5x}{2}} = 1, \quad \lim_{x \to 0} \frac{\sin^2 \frac{5x}{2}}{\left(\frac{5x}{2} \right)^2} = 1. \]

Thus,

\[ \lim_{x \to 0} \frac{\ln(\cos 5x)}{x^2} = 1 \cdot \left( -2 \cdot \frac{25}{4} \cdot 1 \right) = -\frac{25}{2}. \]

L'Hôpital's Rule

Let \( f(x) \) and \( g(x) \) be differentiable in an open interval containing \( a \) (except possibly at \( a \)) and suppose that

\[ \lim_{x \to a} f(x) = 0, \quad \lim_{x \to a} g(x) = 0 \]

or

\[ \lim_{x \to a} f(x) = \pm\infty, \quad \lim_{x \to a} g(x) = \pm\infty. \]

If \( g'(x) \neq 0 \) in this interval and

\[ \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

is a finite real number or \( \pm\infty \), then

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}. \]

This theorem allows us to evaluate certain indeterminate limits by differentiating the numerator and denominator separately. However, its application requires caution.

The differentiability condition ensures that \( f'(x) \) and \( g'(x) \) are well-defined in a neighborhood of \( a \), allowing their ratio to be considered. However, this requires the existence of an open interval around \( a \) where the rule can be applied. If \( g'(x) \) oscillates infinitely often between zero and nonzero values as \( x \to a \), then the necessary interval may not exist, and the application of L’Hôpital’s Rule becomes invalid.

Additionally, the requirement that

\[ \lim_{x \to a} \frac{f'(x)}{g'(x)} \]

be a finite real number or \( \pm\infty \) is essential. If this limit does not settle into a definite value, then L’Hôpital’s Rule does not help in evaluating \( \lim_{x \to a} \frac{f(x)}{g(x)} \). The rule provides a method for computing limits but must be applied with care, ensuring that all conditions are satisfied before using it.

L'Hôpital's Rule for Limits at Infinity

L'Hôpital's Rule also applies to limits as \( x \to \infty \) or \( x \to -\infty \) when dealing with indeterminate forms of type \( \frac{\infty}{\infty} \) or \( \frac{0}{0} \). The theorem is stated as follows:

Let \( f(x) \) and \( g(x) \) be differentiable for all sufficiently large (or small) \( x \), and suppose that

\[ \lim_{x \to \infty} f(x) = \pm\infty, \quad \lim_{x \to \infty} g(x) = \pm\infty \]

or

\[ \lim_{x \to \infty} f(x) = 0, \quad \lim_{x \to \infty} g(x) = 0. \]

If \( g'(x) \neq 0 \) for sufficiently large \( x \) and

\[ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \]

is a finite real number or \( \pm\infty \), then

\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)}. \]

The same result holds for limits as \( x \to -\infty \) under the same conditions.

Similar to previous rule, \( g'(x) \) should not oscillate infinitely often between zero and nonzero values as \( x \to \infty \), since such behavior would prevent a well-defined application of L'Hôpital’s Rule.

Furthermore, the condition that

\[ \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \]

is a finite real number or \( \pm\infty \) is necessary. If this limit does not settle into a well-defined value, then L’Hôpital’s Rule does not yield a conclusion. We cannot conclude anything about the limit in such a case, whether it exists or not.

Consider the limit

\[ \lim_{x \to \infty} \frac{x + \sin x}{x + 1}. \]

Both the numerator and denominator tend to \( \infty \) as \( x \to \infty \), leading to the indeterminate form \( \frac{\infty}{\infty} \). This suggests that L’Hôpital’s Rule might be applicable. Differentiating the numerator and denominator separately, we obtain

\[ \frac{d}{dx} (x + \sin x) = 1 + \cos x, \quad \frac{d}{dx} (x + 1) = 1. \]

Thus, applying L’Hôpital’s Rule formally,

\[ \lim_{x \to \infty} \frac{x + \sin x}{x + 1} = \lim_{x \to \infty} \frac{1 + \cos x}{1} = \lim_{x \to \infty} (1 + \cos x). \]

However, \( \cos x \) oscillates between \( -1 \) and \( 1 \), which means \( 1 + \cos x \) oscillates between \( 0 \) and \( 2 \). Since this expression does not approach a single value, the limit does not exist, making L’Hôpital’s Rule inconclusive.

On the other hand, a direct approach reveals the actual limit. Rewriting the fraction,

\[ \frac{x + \sin x}{x + 1} = \frac{x(1 + \sin x/x)}{x(1 + 1/x)}, \]

Since \( \sin x/x \to 0 \) and \( 1/x \to 0 \) as \( x \to \infty \), this simplifies to

\[ \frac{1 + 0}{1 + 0} = 1. \]

Thus, despite the failure of L'Hôpital's Rule to provide a conclusion, we find that

\[ \lim_{x \to \infty} \frac{x + \sin x}{x + 1} = 1. \]

Repeated Application of L'Hôpital's Rule

L'Hôpital’s Rule can be applied multiple times, provided that all conditions are satisfied at each subsequent application. If, after differentiating once, the resulting limit remains in an indeterminate form, we may differentiate again and continue this process until a determinate form is reached. However, at each step, it must be verified that the quotient of derivatives still satisfies the necessary conditions and that the resulting limit is well-defined as a finite real number or \( \pm\infty \).

Consider the limit

\[ \lim_{x \to 0} \frac{e^x - x - 1}{x^2}. \]

Step 1: Checking the Indeterminate Form

Substituting \( x = 0 \), the numerator becomes

\[ e^0 - 0 - 1 = 1 - 0 - 1 = 0. \]

The denominator is also \( 0^2 = 0 \). Since both the numerator and denominator tend to zero, the given limit is of the indeterminate form \( \frac{0}{0} \), allowing us to apply L’Hôpital’s Rule.

Step 2: First Application of L'Hôpital's Rule

Differentiating the numerator and denominator separately,

\[ \frac{d}{dx} (e^x - x - 1) = e^x - 1, \quad \frac{d}{dx} (x^2) = 2x. \]

Thus, applying L’Hôpital’s Rule,

\[ \lim_{x \to 0} \frac{e^x - x - 1}{x^2} = \lim_{x \to 0} \frac{e^x - 1}{2x}. \]

Substituting \( x = 0 \) again, the numerator is \( e^0 - 1 = 0 \), and the denominator is \( 2(0) = 0 \), giving another \( \frac{0}{0} \) form. Since the indeterminate form persists, we apply L'Hôpital’s Rule once more.

Step 3: Second Application of L'Hôpital's Rule

Differentiating again,

\[ \frac{d}{dx} (e^x - 1) = e^x, \quad \frac{d}{dx} (2x) = 2. \]

Thus,

\[ \lim_{x \to 0} \frac{e^x - 1}{2x} = \lim_{x \to 0} \frac{e^x}{2}. \]

Now, substituting \( x = 0 \), we obtain

\[ \frac{e^0}{2} = \frac{1}{2}. \]

Thus, the given limit evaluates to

\[ \lim_{x \to 0} \frac{e^x - x - 1}{x^2} = \frac{1}{2}. \]

This example illustrates how L'Hôpital’s Rule can be applied multiple times if necessary, provided that the conditions remain valid at each step. The process continues until a determinate limit is reached.

Avoiding Unnecessary Complexity in L'Hôpital's Rule

While L'Hôpital’s Rule is a powerful tool for evaluating limits, applying it blindly can sometimes lead to an increase in complexity rather than simplifying the problem. In many cases, differentiating both the numerator and denominator makes the expressions more complicated without immediately resolving the indeterminate form. If the resulting limit remains indeterminate, one might be tempted to apply L’Hôpital’s Rule again, but this can further aggravate the complexity, making the evaluation unnecessarily cumbersome.

To prevent this, here are some strategies for reducing complexity before differentiating:

1. Simplify the expression before applying L’Hôpital’s Rule –

   A preliminary algebraic simplification may result in more manageable derivatives.

2. Identify and remove non-contributing factors –

   If a term in the numerator or denominator is in product with the rest but does not contribute to the indeterminate form (i.e., its own limit is finite and nonzero), then it can be removed by directly applying its limit.

For example, consider the following limit:

\[ \lim_{x \to 0} \frac{(\sin x + \cos x)(1 - \cos x)}{x^2}. \]

Step 1: Identifying Non-Contributing Factors

As \( x \to 0 \),

\[ \sin x + \cos x \to 1. \]

Since this factor does not contribute to the \( 0/0 \) form, it can be removed by applying its limit separately. This reduces the original limit to

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2}. \]

Step 2: Applying L’Hôpital’s Rule

The numerator and denominator both tend to 0, so we apply L'Hôpital’s Rule:

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\sin x}{2x}. \]

Again, this is of the form \( \frac{0}{0} \), so we apply L'Hôpital’s Rule once more:

\[ \lim_{x \to 0} \frac{\cos x}{2} = \frac{\cos 0}{2} = \frac{1}{2}. \]

Thus,

\[ \lim_{x \to 0} \frac{(\sin x + \cos x)(1 - \cos x)}{x^2} = \frac{1}{2}. \]

Conclusion

Had we applied L'Hôpital’s Rule directly to the original fraction, differentiating the entire numerator and denominator would have resulted in more complex expressions. Instead, recognizing and removing the non-contributing factor simplified the expression significantly before differentiation.

Thus, before applying L’Hôpital’s Rule, always check if there are terms that can be simplified or removed by applying their limits. This approach can prevent unnecessary differentiation and reduce computational complexity.