Evaluation of Limits
Infinite Limits
An infinite limit describes the behavior of a function \( f(x) \) as \( x \) approaches a certain value \( a \), where \( f(x) \) grows arbitrarily large (positively or negatively). For instance, if \( f(x) \to \infty \) as \( x \to a \), we write:
This notation signifies that the values of \( f(x) \) increase without bound as \( x \) gets closer and closer to \( a \). Similarly, if \( f(x) \to -\infty \), we write:
indicating that the function decreases without bound as \( x \) approaches \( a \).
To better understand this concept, consider the function \( f(x) = \frac{1}{x} \).
Example
Evaluate: \( \lim_{x \to 0^+} \frac{1}{x} \) and \( \lim_{x \to 0^-} \frac{1}{x} \)
Let us evaluate the behavior of \( f(x) = \frac{1}{x} \) as \( x \) approaches \( 0 \) from the positive and negative sides. We start by approaching \( 0 \) from the right (\( x > 0 \)):
As \( x \to 0^+ \), the values of \( f(x) \) grow larger and larger. Hence, we write:
Next, let us approach \( 0 \) from the left (\( x < 0 \)):
As \( x \to 0^- \), the values of \( f(x) \) decrease without bound. Hence, we write:
The results \( \lim_{x \to 0^+} \frac{1}{x} = \infty \) and \( \lim_{x \to 0^-} \frac{1}{x} = -\infty \) describe how the function behaves near \( x = 0 \). Although the limit \( \lim_{x \to 0} \frac{1}{x} \) does not exist in the conventional sense, specifying \( \infty \) or \( -\infty \) gives us a clear picture of the unbounded growth of the function in either direction.
Example
Prove that: \( \lim_{x \to 0^+} \ln x = -\infty \)
Consider the function \( f(x) = \ln x \). We want to explore the behavior of \( f(x) \) as \( x \to 0^+ \) (i.e., \( x \) approaches \( 0 \) from the positive side). Let us evaluate \( f(x) \) for a sequence of values \( x = \frac{1}{e}, \frac{1}{e^2}, \frac{1}{e^3}, \dots \), where each term becomes progressively smaller and closer to \( 0 \).
As \( x \to 0^+ \), the values of \( \ln x \) decrease without bound. Hence, we write:
This conclusion can be further verified by examining the graph of \( f(x) = \ln x \), which we already know from basic function properties. The logarithmic function is defined for \( x > 0 \), and as \( x \) approaches \( 0^+ \), the graph of \( \ln x \) descends steeply without bound, confirming that \( \ln x \to -\infty \).
The graph provides an intuitive understanding of the function's behavior, making it clear why \( \ln x \) decreases indefinitely as \( x \) approaches \( 0^+ \). This is why learning the shapes and behaviors of standard graphs is essential—they allow us to make such conclusions with confidence, even without explicit numerical calculations.
The graph of \( f(x) = \ln x \) visually demonstrates the following key behaviors:
- Unbounded decrease as \( x \to 0^+ \): The curve descends steeply toward negative infinity, confirming that \( \ln x \to -\infty \) as \( x \to 0^+ \).
- Slow growth for large \( x \): The logarithmic curve rises slowly for large values of \( x \), which is why \( \ln x \to \infty \) as \( x \to \infty \).
Thus, graphs of fundamental functions like \( \ln x \), \( e^x \), and \( x^n \) are invaluable tools in understanding limits, especially infinite limits. By referring to these graphs, we can immediately recognize behaviors like rapid unbounded descent, which would otherwise require detailed numerical verification. This reinforces the importance of familiarizing ourselves with the key features of standard graphs in calculus.
Example
Evaluate:
- \( \lim_{x \to \frac{\pi}{2}^-} \tan x \)
- \( \lim_{x \to \frac{\pi}{2}^+} \tan x \)
Solution:
The function \( \tan x \) is defined as \( \tan x = \frac{\sin x}{\cos x} \). To evaluate the given limits, we analyze the behavior of \( \sin x \) and \( \cos x \) as \( x \) approaches \( \frac{\pi}{2} \) from the left and right.
-
Evaluate \( \lim_{x \to \frac{\pi}{2}^-} \tan x \):
As \( x \to \frac{\pi}{2}^- \), \( x \) is in the first quadrant where:
- \( \sin x > 0 \) and approaches \( 1 \) as \( x \to \frac{\pi}{2}^- \).
- \( \cos x > 0 \) but decreases toward \( 0 \) as \( x \to \frac{\pi}{2}^- \).
Since \( \tan x = \frac{\sin x}{\cos x} \) and the numerator \( \sin x \to 1 \) while the denominator \( \cos x \to 0^+ \) (positive but very small), the fraction grows arbitrarily large. Hence:
\[ \lim_{x \to \frac{\pi}{2}^-} \tan x = +\infty. \] -
Evaluate \( \lim_{x \to \frac{\pi}{2}^+} \tan x \):
As \( x \to \frac{\pi}{2}^+ \), \( x \) is in the second quadrant where:
- \( \sin x > 0 \) and approaches \( 1 \) as \( x \to \frac{\pi}{2}^+ \).
- \( \cos x < 0 \) and increases toward \( 0 \) from the negative side as \( x \to \frac{\pi}{2}^+ \).
Since \( \tan x = \frac{\sin x}{\cos x} \) and the numerator \( \sin x \to 1 \) while the denominator \( \cos x \to 0^- \) (negative but very small), the fraction decreases without bound. Hence:
\[ \lim_{x \to \frac{\pi}{2}^+} \tan x = -\infty. \]
Refer to the graph of \( \tan x \), which has vertical asymptotes at \( x = \frac{\pi}{2} + n\pi \) for integers \( n \). The graph confirms:
- \( \tan x \to +\infty \) as \( x \to \frac{\pi}{2}^- \),
- \( \tan x \to -\infty \) as \( x \to \frac{\pi}{2}^+ \).
This graphical behavior corroborates the results derived analytically.
General form of infinite limits
When evaluating limits of the form
where \( f(x) \to c \) (\( c \neq 0 \)) and \( g(x) \to 0 \), the result is an infinite limit. The limit diverges to \( +\infty \) or \( -\infty \) depending on the signs of \( f(x) \) and \( g(x) \) as \( x \to a \).
Consider the example:
Here, \( f(x) = x+1 \) and \( g(x) = x-1 \). As \( x \to 1^+ \), \( x+1 \to 2 \), which is a finite positive constant. Simultaneously, \( x-1 \to 0^+ \) because \( x > 1 \) near \( 1 \), making \( g(x) \) a small positive value approaching \( 0 \). Since the numerator remains positive and the denominator approaches \( 0^+ \), the fraction grows arbitrarily large in the positive direction. Thus:
The positive sign of \( g(x) \) near \( x \to 1^+ \) is crucial in determining the behavior of the limit.
Limits at Infinity
Consider a function \( f(x) \) defined for \( x > a \), where \( a \) is some real number. We are interested in analyzing the behavior of \( f(x) \) as \( x \to \infty \), that is, as \( x \) becomes arbitrarily large. If \( f(x) \) approaches a fixed real number \( L \) as \( x \to \infty \), we write:
This statement means that, no matter how large \( x \) becomes, the values of \( f(x) \) get arbitrarily close to \( L \) and remain near \( L \). Similarly, if \( f(x) \) is defined for \( x < a \) and \( f(x) \) approaches a fixed real number \( L \) as \( x \to -\infty \), we write:
If \( f(x) \) does not approach a finite value but instead grows without bound as \( x \to \infty \) or \( x \to -\infty \), the limit is said to diverge. In this case, we write:
Example
Evaluate: \( \lim_{x \to \infty} \frac{1}{x} \)
**Solution: **
As \( x \to \infty \), the numerator \( 1 \) remains constant, while the denominator \( x \) grows arbitrarily large. Dividing a fixed number by an increasingly large quantity results in a value that gets closer and closer to \( 0 \). Therefore:
This conclusion is consistent with the observation that the larger the value of \( x \), the smaller the value of \( \frac{1}{x} \), approaching \( 0 \) but never actually reaching it.
Divergent Behavior
Evaluate: \( \lim_{x \to \infty} x^2 \)
Solution:
As \( x \to \infty \), the value of \( x^2 \) grows arbitrarily large. There is no finite number that \( f(x) \) approaches as \( x \to \infty \). Instead, \( f(x) \) increases without bound. Hence, the limit diverges, and we write:
Similarly, as \( x \to -\infty \), \( x^2 \) also grows arbitrarily large because squaring a negative number results in a positive value. Therefore:
Evaluating Limits at Infinity using Reciprocal Substitution
To simplify the evaluation of limits at infinity, the variable \( x \) can be replaced with another variable \( y \), defined as \( y = \frac{1}{x} \). With this substitution, the behavior of \( x \) as it approaches \( \infty \) or \( -\infty \) corresponds to the behavior of \( y \) near \( 0 \). Specifically:
- As \( x \to \infty \), \( y \to 0^+ \) (a small positive number).
- As \( x \to -\infty \), \( y \to 0^- \) (a small negative number).
This substitution is particularly useful in transforming complicated expressions involving large values of \( x \) into simpler forms that are easier to analyze as \( y \to 0 \).
Limits of Exponential Functions at Infinity and Negative Infinity
The behavior of the exponential function \( a^x \) depends on the value of the base \( a \) and the direction in which \( x \) approaches infinity or negative infinity.
Case 1: \( 0 < a < 1 \), as \( x \to \infty \)
When \( 0 < a < 1 \), the base \( a \) is a fraction, and each successive power of \( a \) reduces the magnitude of \( a^x \). This happens because multiplying by a fraction repeatedly diminishes the result. As \( x \to \infty \), the values of \( a^x \) approach \( 0 \).
To justify, let us consider \( a = \frac{1}{2} \). The table below shows the values of \( \left(\frac{1}{2}\right)^x \) as \( x \to \infty \):
From the table, we observe that as \( x \to \infty \), the successive values of \( a^x \) get closer and closer to \( 0 \), though they never actually reach \( 0 \). This is a defining characteristic of exponential decay when \( 0 < a < 1 \). Therefore:
Case 2: \( a > 1 \), as \( x \to \infty \)
When \( a > 1 \), the base \( a \) is greater than 1, and each successive power of \( a \) increases the magnitude of \( a^x \). As \( x \to \infty \), the values of \( a^x \) grow without bound, leading to divergence to infinity.
To justify, let us consider \( a = 2 \). The table below shows the values of \( 2^x \) as \( x \to \infty \):
From the table, it is clear that as \( x \to \infty \), \( 2^x \) increases without bound. This demonstrates the unbounded growth of exponential functions when \( a > 1 \). Hence:
Case 3: \( 0 < a < 1 \), as \( x \to -\infty \)
When \( x \to -\infty \), the exponent \( x \) is negative, and \( a^x = \frac{1}{a^{-x}} \). Since \( 0 < a < 1 \), its reciprocal \( \frac{1}{a} > 1 \). As \( x \to -\infty \), \( a^{-x} \to \infty \), making \( \frac{1}{a^{-x}} \to \infty \). Thus, \( a^x \) grows without bound as \( x \to -\infty \).
To justify, let us consider \( a = \frac{1}{2} \). The table below shows the values of \( \left(\frac{1}{2}\right)^x \) as \( x \to -\infty \):
From the table, we observe that as \( x \to -\infty \), \( a^x \) grows arbitrarily large. Hence:
Case 4: \( a > 1 \), as \( x \to -\infty \)
When \( x \to -\infty \), the exponent \( x \) is negative, so \( a^x = \frac{1}{a^{-x}} \). Since \( a > 1 \), \( a^{-x} \to \infty \) as \( x \to -\infty \), making \( \frac{1}{a^{-x}} \to 0 \). Therefore, \( a^x \) approaches \( 0 \) as \( x \to -\infty \).
To justify, let us consider \( a = 2 \). The table below shows the values of \( 2^x \) as \( x \to -\infty \):
From the table, it is clear that as \( x \to -\infty \), \( a^x \) gets closer and closer to \( 0 \), though it never actually reaches \( 0 \). Hence:
Hence,
Limits of Polynomials at Infinity
We want to understand how a polynomial function behaves as \(x\) becomes very large (\(x \to \infty\)) or very small (\(x \to -\infty\)). Consider the general polynomial:
where \(a_0 \neq 0\) and \(n \in \mathbb{N}\). The term \(a_0x^n\) is called the leading term, and \(a_0\) is its leading coefficient. The leading term is crucial in determining the behavior of \(p(x)\) because it dominates all other terms for large values of \(|x|\).
To see this, let us rewrite \(p(x)\) as:
Notice that as \(x \to \infty\) or \(x \to -\infty\), the terms \(\frac{a_1}{a_0x}, \frac{a_2}{a_0x^2}, \dots, \frac{a_n}{a_0x^n}\) all approach \(0\), since the denominators grow without bound. Therefore, \(p(x)\) behaves like its leading term \(a_0x^n\), and the contributions of the remaining terms become negligible.
Now, we analyze the behavior of \(p(x)\) by considering the degree \(n\) (even or odd) and the sign of the leading coefficient \(a_0\).
Case 1: \(n\) is even For an even degree \(n\), the power \(x^n\) is always positive regardless of whether \(x > 0\) or \(x < 0\). This simplifies the analysis as follows:
Subcase 1: \(a_0 > 0\)
- When \(x \to \infty\), \(x^n \to \infty\), and since \(a_0 > 0\), the leading term \(a_0x^n \to \infty\). Hence, \(p(x) \to \infty\).
- When \(x \to -\infty\), \(x^n > 0\) (as the even power of a negative number is positive), and \(a_0x^n \to \infty\). Hence, \(p(x) \to \infty\).
In this case, the polynomial diverges to \(+\infty\) in both directions.
Subcase 2: \(a_0 < 0\)
- When \(x \to \infty\), \(x^n \to \infty\), but since \(a_0 < 0\), the leading term \(a_0x^n \to -\infty\). Hence, \(p(x) \to -\infty\).
- When \(x \to -\infty\), \(x^n > 0\), and \(a_0x^n \to -\infty\). Hence, \(p(x) \to -\infty\).
In this case, the polynomial diverges to \(-\infty\) in both directions.
Thus,
- If \(a_0 > 0\), \(p(x) \to +\infty\) as \(x \to \pm\infty\).
- If \(a_0 < 0\), \(p(x) \to -\infty\) as \(x \to \pm\infty\).
Case 2: \(n\) is odd
For an odd degree \(n\), the power \(x^n\) has the same sign as \(x\). Specifically, \(x^n > 0\) for \(x > 0\) and \(x^n < 0\) for \(x < 0\). Let us analyze:
Subcase 1: \(a_0 > 0\)
- When \(x \to \infty\), \(x^n \to \infty\), and since \(a_0 > 0\), the leading term \(a_0x^n \to \infty\). Hence, \(p(x) \to \infty\).
- When \(x \to -\infty\), \(x^n \to -\infty\) (as the odd power of a negative number is negative), and \(a_0x^n \to -\infty\). Hence, \(p(x) \to -\infty\).
In this case, the polynomial rises to \(+\infty\) as \(x \to \infty\) and falls to \(-\infty\) as \(x \to -\infty\).
Subcase 2: \(a_0 < 0\)
- When \(x \to \infty\), \(x^n \to \infty\), but since \(a_0 < 0\), the leading term \(a_0x^n \to -\infty\). Hence, \(p(x) \to -\infty\).
- When \(x \to -\infty\), \(x^n \to -\infty\), and \(a_0x^n \to \infty\) (as the product of two negative numbers is positive). Hence, \(p(x) \to \infty\).
In this case, the polynomial falls to \(-\infty\) as \(x \to \infty\) and rises to \(+\infty\) as \(x \to -\infty\).
Hence,
- If \(a_0 > 0\), \(p(x) \to +\infty\) as \(x \to \infty\) and \(p(x) \to -\infty\) as \(x \to -\infty\).
- If \(a_0 < 0\), \(p(x) \to -\infty\) as \(x \to \infty\) and \(p(x) \to +\infty\) as \(x \to -\infty\).
The behavior of a polynomial function \(p(x)\) at infinity and negative infinity is entirely governed by its leading term \(a_0x^n\). The degree \(n\) (even or odd) determines whether the polynomial behaves symmetrically or asymmetrically about the \(x\)-axis, while the sign of \(a_0\) dictates whether it rises to \(+\infty\) or falls to \(-\infty\).
Thus:
-
For \(n\) even:
- \(a_0 > 0\): \(p(x) \to +\infty\) as \(x \to \pm\infty\).
- \(a_0 < 0\): \(p(x) \to -\infty\) as \(x \to \pm\infty\).
-
For \(n\) odd:
- \(a_0 > 0\): \(p(x) \to +\infty\) as \(x \to \infty\), and \(p(x) \to -\infty\) as \(x \to -\infty\).
- \(a_0 < 0\): \(p(x) \to -\infty\) as \(x \to \infty\), and \(p(x) \to +\infty\) as \(x \to -\infty\).
Example
I.
We can see this behavior easily from the graph of \(y = \tan^{-1}x\). As \(x \to \infty\), the function approaches the horizontal asymptote \(y = \frac{\pi}{2}\), and as \(x \to -\infty\), it approaches the horizontal asymptote \(y = -\frac{\pi}{2}\).
II.
We can observe this behavior clearly from the graph of \(y = \cot^{-1}x\). As \(x \to \infty\), the function approaches the horizontal asymptote \(y = 0\), and as \(x \to -\infty\), it approaches the horizontal asymptote \(y = \pi\).
III.
let us analyze the behavior of the numerator and the denominator separately.
-
Behavior of the numerator, \(\sin x\):
The function \(\sin x\) is periodic, oscillating between \(-1\) and \(1\), regardless of the value of \(x\). That is, \(|\sin x| \leq 1\) for all \(x\). The numerator does not grow unbounded; instead, it is restricted to this fixed range.
-
Behavior of the denominator, \(x\):
As \(x \to \infty\), the denominator \(x\) increases without bound. Unlike the numerator, the denominator grows arbitrarily large.
When these two components are combined in the fraction \(\frac{\sin x}{x}\), the bounded nature of the numerator (\(\sin x\)) is overwhelmed by the unbounded growth of the denominator (\(x\)).
To make this more concrete, consider evaluating the fraction for increasingly larger values of \(x\). For example:
As \(x\) increases, regardless of the oscillations of \(\sin x\), the fraction \(\frac{\sin x}{x}\) becomes smaller and smaller, approaching \(0\).
Thus, the limit of \(\frac{\sin x}{x}\) as \(x \to \infty\) is:
Sandwich Theorem
The Sandwich Theorem (also called the Squeeze Theorem) is a fundamental result in calculus that helps evaluate the limit of a function by "sandwiching" it between two other functions whose limits are known and equal.
Theorem Statement
Let \( f(x) \), \( g(x) \), and \( h(x) \) be functions defined on an interval \( I \), except possibly at a point \( c \), where \( c \) is a limit point of \( I \). Suppose that:
- \( g(x) \leq f(x) \leq h(x) \) for all \( x \) in \( I \) and near \( c \),
- \( \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L \), where \(L \in \mathbb{R}\) or \(L = \infty\) or \(L = -\infty\)
Then:
It can be understood intuitively as follows:
Suppose \( f(x) \) is a function you're interested in studying. Imagine it as a curve whose behavior near a point \( c \) (or as \( x \to c \)) is uncertain. To understand what \( f(x) \) is doing, you cleverly find two other functions, \( g(x) \) and \( h(x) \), that "sandwich" \( f(x) \). This means that near \( c \), the inequality holds:
The catch is that these two "sandwiching" functions \( g(x) \) and \( h(x) \) must satisfy an important condition: both must have the same limit \( L \) as \( x \to c \). That is:
where \( L \) can be any real number (\( L \in \mathbb{R} \)), or even \( +\infty \) or \( -\infty \).
If this is true, then the function \( f(x) \), being "trapped" between \( g(x) \) and \( h(x) \), is also forced to have the same limit \( L \). Symbolically:
In simpler terms: if you can tightly squeeze \( f(x) \) between two functions that both converge to the same value, then \( f(x) \) must also converge to that value. The logic here is that \( f(x) \) has no room to "escape" from \( L \).
Example
Evaluate:
Solution:
-
Let \( f(x) = x^2 \sin\left(\frac{1}{x}\right) \). The sine function satisfies \( -1 \leq \sin\left(\frac{1}{x}\right) \leq 1 \).
-
Multiplying through by \( x^2 \) (which is non-negative for \( x > 0 \)), we get:
\[ -x^2 \leq x^2 \sin\left(\frac{1}{x}\right) \leq x^2. \] -
Consider the functions \( g(x) = -x^2 \) and \( h(x) = x^2 \). Clearly:
\[ \lim_{x \to 0} g(x) = \lim_{x \to 0} h(x) = 0. \] -
By the Sandwich Theorem:
\[ \lim_{x \to 0} x^2 \sin\left(\frac{1}{x}\right) = 0. \]
Sandwich Theorem for Limit at Infinity
Let \( f(x) \), \( g(x) \), and \( h(x) \) be functions defined on the interval \((a, \infty)\). Suppose that:
- \( g(x) \leq f(x) \leq h(x) \) for all \( x > a \),
- \( \lim_{x \to \infty} g(x) = \lim_{x \to \infty} h(x) = L \), where \( L \) is a real number (\( L \in \mathbb{R} \)) or \( \pm\infty \).
Then:
If \( f(x) \) is "trapped" between two functions, \( g(x) \) and \( h(x) \), and these two bounding functions both converge to the same limit \( L \) as \( x \to \infty \), then \( f(x) \) is forced to converge to \( L \). The oscillations or variations in \( f(x) \) become irrelevant because it has no room to "escape" from the bounds set by \( g(x) \) and \( h(x) \).
Example
Evaluate:
Solution:
Let \( f(x) = \frac{\sin x}{x} \). Since \( \sin x \) oscillates between \(-1\) and \( 1 \), we have:
Here, the bounding functions are \( g(x) = -\frac{1}{x} \) and \( h(x) = \frac{1}{x} \). As \( x \to \infty \), both \( g(x) \) and \( h(x) \) converge to \( 0 \):
By the Sandwich Theorem:
Thus, the limit of \( f(x) = \frac{\sin x}{x} \) as \( x \to \infty \) is \( 0 \).
Example
Evaluate:
where \(\lfloor x \rfloor\) is the greatest integer function (floor function).
Solution:
To solve, recall the fundamental property of the greatest integer function:
Dividing through by \(x > 0\), we get:
Simplifying the bounds:
As \(x \to \infty\), observe:
Thus, by the Sandwich Theorem, the limit of \(\frac{\lfloor x \rfloor}{x}\) is:
The limit is:
Determinate Forms of Limits
To understand what determinate forms of limits are, consider the following examples.
Example 1: \(\lim_{x \to 0^+} \frac{x+1}{x}\)
Examine the function \(f(x) = \frac{x+1}{x}\) as \(x\) approaches \(0^+\).
The numerator, \(x+1\), simplifies to \(1\) because \(x \to 0^+\) implies \(x\) becomes negligibly small, leaving \(1\) unchanged. Thus,
The denominator, \(x\), approaches \(0\) from the positive side since \(x > 0\). Therefore,
Substituting these behaviors into the expression, we see that \(f(x)\) takes the form \(\frac{1}{0^+}\). This form signifies that the fraction grows unbounded as \(x\) approaches \(0^+\). Thus,
Example 2: \(\lim_{x \to 0^+} \frac{x^2+1}{x^3}\)
Now consider the function \(f(x) = \frac{x^2+1}{x^3}\). Let us analyze its behavior as \(x \to 0^+\).
The numerator, \(x^2+1\), reduces to \(1\), as the term \(x^2\) vanishes when \(x\) becomes very small. Therefore,
The denominator, \(x^3\), shrinks to \(0^+\), as \(x^3\) approaches \(0\) while remaining positive. Hence,
This again results in the form \(\frac{1}{0^+}\), which is determinate and leads to \(+\infty\). Consequently,
The form \(\frac{1}{0^+}\) is known as a determinate form because its outcome is always \(+\infty\), irrespective of the exact expressions for the numerator and denominator, provided:
- The numerator approaches a positive finite value.
- The denominator approaches \(0\) from the positive side.
If we take another example like
we observe the following.
As \(x \to 0^+\), the numerator, \(1 + 2\sin^3x\), simplifies to \(1\) since \(\sin^3x\) becomes negligibly small as \(\sin x \to 0^+\). Hence,
The denominator, \(\sin x\), approaches \(0^+\), as \(\sin x > 0\) for \(x > 0\) and \(x \to 0^+\). Thus,
This results in the determinate form \(\frac{1}{0^+}\), indicating that the value of the limit grows without bound. Therefore,
The outcome once again confirms that the form \(\frac{1}{0^+}\) always leads to \(+\infty\), irrespective of the exact nature of the functions in the numerator and denominator, as long as their behavior satisfies the stated conditions.
A determinate form of a limit is one where the value of the limit does not depend on the specific functions involved but solely on the limiting behavior of the terms involved in the expression. Let us take another example.
Example
Consider the evaluation of
To analyze this expression, we first examine the behavior of the base and the power as \(x \to 1^+\).
The base, \(\frac{1}{x+1}\), approaches \(\frac{1}{2}\). This is because as \(x \to 1^+\), the term \(x+1\) approaches \(2\). Hence, the base remains finite and converges to a fixed value, \(\frac{1}{2}\).
The power, \(\frac{1}{x-1}\), diverges to \(+\infty\). To see why, note that as \(x \to 1^+\), the term \(x-1\) becomes arbitrarily small and positive (\(x > 1\)). This results in the form \(\frac{1}{0^+}\), which indicates that the power grows without bound.
Thus, the expression as \(x \to 1^+\) takes the form \(\left(\frac{1}{2}\right)^{+\infty}\). A base of \(\frac{1}{2}\) raised to an infinitely large power approaches \(0\), because repeated multiplication of a fraction less than \(1\) shrinks it indefinitely. Therefore,
Consider a different expression:
Here too, we analyze the behavior of the base and the power as \(x \to \frac{\pi}{3}^+\).
The base, \(\cos x\), approaches \(\frac{1}{2}\). As \(x\) nears \(\frac{\pi}{3}\) from the right, \(\cos x\) converges to \(\frac{1}{2}\), since \(\cos(\frac{\pi}{3}) = \frac{1}{2}\).
The power, \(\cot(x - \frac{\pi}{3})\), diverges to \(+\infty\). This is because \(x - \frac{\pi}{3} \to 0^+\) as \(x \to \frac{\pi}{3}^+\), making \(\cot(x - \frac{\pi}{3}) = \frac{\cos(x - \frac{\pi}{3})}{\sin(x - \frac{\pi}{3})}\) behave like \(\frac{1}{0^+}\), which grows unboundedly.
Thus, the expression takes the form \((\frac{1}{2})^{+\infty}\), which, as in the previous example, evaluates to \(0\). Therefore,
In both cases, the base approaches \(\frac{1}{2}\), and the power diverges to \(+\infty\). The resulting form, \((\frac{1}{2})^{+\infty}\), consistently leads to a limit of \(0\). Thus it is determinate.
Common Determinate Forms of Limits
The following are determinate forms of limits. These arise when evaluating expressions where the involved quantities approach specific values. In these forms, \(\alpha \in \mathbb{R}\) represents a finite value to which a limit is approaching, while \(\infty\) or \(-\infty\) indicates divergence to positive or negative infinity.
These forms always yield well-defined conclusions, ensuring clarity in handling such limits.
-
Addition of Infinity:
- \(\alpha + \infty = \infty\), where \(\alpha\) is a finite number.
- \(\alpha - \infty = -\infty\), where \(\alpha\) is a finite number.
-
Multiplication with Infinity:
- \(\alpha \cdot \infty = \infty\), if \(\alpha > 0\).
- \(\alpha \cdot \infty = -\infty\), if \(\alpha < 0\).
- \(\infty \cdot \infty = \infty\).
-
Addition of Infinities:
- \(\infty + \infty = \infty\).
-
Division by Zero (One-Sided Limits):
- \(\frac{\alpha}{0^+} = \infty\), if \(\alpha > 0\).
- \(\frac{\alpha}{0^+} = -\infty\), if \(\alpha < 0\).
- \(\frac{\alpha}{0^-} = -\infty\), if \(\alpha > 0\).
- \(\frac{\alpha}{0^-} = \infty\), if \(\alpha < 0\).
-
Powers of Zero and Infinity:
- \(0^{\infty} = 0\), where the base is approaching \(0\), not exactly \(0\).
- \(\infty^{\infty} = \infty\).
Please note:
- These are determinate because they always lead to unique outcomes. For example, the result of \(\frac{\alpha}{0^+}\) is entirely determined by the sign of \(\alpha\) and the direction of approach of the denominator.
- These forms do not involve ambiguities like those found in indeterminate forms such as \(\frac{0}{0}\), \(\infty - \infty\), or \(0 \cdot \infty\).
- It is crucial to carefully interpret the behavior of the involved terms, especially when one-sided limits or exponents are present, to apply these forms correctly.
These forms clarify how limits behave in straightforward cases, ensuring there is no confusion when encountering such expressions.
Indeterminate Forms of Limits
In contrast to determinate forms, we encounter a very special category of limits known as indeterminate forms. These are situations where the limiting behavior of the terms involved does not uniquely determine the value of the limit. Instead, the result depends on the specific nature of the functions involved in the expression. Let us understand this through illustrative examples.
Example 1: \(\lim_{x \to 0} \frac{x}{x}\)
Consider the limit \(\frac{x}{x}\) as \(x \to 0\). At first glance, the numerator and denominator both approach \(0\), leading to the form \(\frac{0}{0}\). However, observing the structure of the fraction, we see that the terms cancel, leaving:
Hence,
Here, although the form is \(\frac{0}{0}\), the specific nature of the numerator and denominator determines the result.
Example 2: \(\lim_{x \to 0} \frac{2x}{x}\)
Now consider the limit \(\frac{2x}{x}\). Again, the form is \(\frac{0}{0}\) as \(x \to 0\), since both numerator and denominator tend to \(0\). Simplifying the fraction yields:
Thus,
Example 3: \(\lim_{x \to 0} \frac{3x}{x}\)
Similarly, for \(\frac{3x}{x}\), we observe the same form, \(\frac{0}{0}\). Simplifying, we find:
Hence,
Example 4: \(\lim_{x \to 0} \frac{x^2}{x}\)
In this case, the form is again \(\frac{0}{0}\), as both the numerator \(x^2\) and the denominator \(x\) approach \(0\). However, simplifying gives:
Thus, as \(x \to 0\), we have:
The examples above reveal that the form \(\frac{0}{0}\) does not uniquely determine the value of the limit. Instead, the result depends entirely on the specific expressions for the numerator and denominator. This makes \(\frac{0}{0}\) an indeterminate form—its value cannot be predicted solely from the form itself.
Apart from \(\frac{0}{0}\), there are six additional indeterminate forms that arise in various contexts. These include:
Each of these forms requires careful analysis of the functions involved to determine the correct value of the limit. These forms, like \(\frac{0}{0}\), are "indeterminate" because the result depends on the interplay between the terms, not merely on the form itself.
We will learn various techniques to deal with these forms. While determinate forms can be handled with straightforward reasoning, it is the *indeterminate forms that demand our primary focus and deeper analysis.***
Algebra of Limits
Let \(f(x)\) and \(g(x)\) be two functions such that
where \(L_f\) and \(L_g\) are finite real numbers. The following theorems provide essential tools for manipulating and combining limits.
-
Constant Multiple Rule:
For any constant \(k\),
\[ \lim_{x \to a} [k \cdot f(x)] = k \cdot \lim_{x \to a} f(x) = k \cdot L_f. \]A constant factor can be taken outside the limit.
-
Sum Rule:
The limit of the sum of two functions is equal to the sum of their limits:
\[ \lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) = L_f + L_g. \] -
Difference Rule:
The limit of the difference of two functions is equal to the difference of their limits:
\[ \lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) = L_f - L_g. \] -
Product Rule:
The limit of the product of two functions is equal to the product of their limits:
\[ \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) = L_f \cdot L_g. \] -
Quotient Rule:
If \(L_g \neq 0\), the limit of the quotient of two functions is equal to the quotient of their limits:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} = \frac{L_f}{L_g}. \] -
Power Rule:
If \(n\) is a positive integer,
\[ \lim_{x \to a} [f(x)]^n = \left(\lim_{x \to a} f(x)\right)^n = L_f^n. \]For rational powers,
\[ \lim_{x \to a} [f(x)]^{m/n} = \left(\lim_{x \to a} f(x)\right)^{m/n} = L_f^{m/n}, \]provided the limit exists and the expression is well-defined.
These theorems are fundamental tools for evaluating limits and will be applied frequently to simplify and compute various expressions involving limits.
Continuous Functions and Evaluating Limits by Direct Subsitution
Intuitively, a function is continuous if its graph has no breaks, jumps, or holes. This means the function behaves predictably, and we can smoothly trace it without lifting our pen. Most functions we encounter in our course are continuous or atleast only not continuous at some points (that is, they are continuous in pieces).
Definition of Continuity
A function \(f(x)\) is said to be continuous at a point \(x = a\) if the following three conditions are satisfied:
- The left-hand limit (LHL) of \(f(x)\) as \(x \to a\) exists.
- The right-hand limit (RHL) of \(f(x)\) as \(x \to a\) exists.
-
The value of the function at \(x = a\), denoted by \(f(a)\), exists, and all three are equal:
\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a). \]
In simple terms, the function has no sudden changes at \(x = a\), and the limit matches the actual value of the function at that point.
Calculating Limits for Continuous Functions via direct substitution
For a continuous function \(f(x)\), if someone asks to calculate the limit at a point \(x = a\), we can directly substitute \(a\) into \(f(x)\) to find the limit. That is:
For example, if \(f(x) = x^2\), which is continuous everywhere, the limit at \(x = 2\) is simply:
When Direct Substitution Fails
There are situations where we cannot use direct substitution to evaluate limits, even if \(f(x)\) seems continuous. These include:
-
Indeterminate Forms:
If substituting \(x = a\) gives forms like \(\frac{0}{0}, 0 \cdot \infty, \infty - \infty\), etc., the limit cannot be directly computed, and additional techniques are required.
-
Points Where Terms Approach Infinity:
If parts of the function approach infinity as \(x \to a\), substitution does not give a meaningful result, and we need to analyze the asymptotic behavior.
-
Discontinuities:
If \(f(x)\) is known to be discontinuous at \(x = a\), substitution cannot be used to find the limit. For instance, the greatest integer function \([x]\) (floor function) is discontinuous at all integers because it "jumps" at these points.
For piece-wise defined functions as well , the substitution may fail at transition points where function changes definition. e.g.
Consider the function:
\[ f(x) = \begin{cases} x + 1, & x < 2, \\ 3x - 4, & x \geq 2. \end{cases} \]At \(x = 2\), the transition point:
-
For \(x \to 2^-\), use \(f(x) = x + 1\):
\[ \lim_{x \to 2^-} f(x) = 2 + 1 = 3. \] -
For \(x \to 2^+\), use \(f(x) = 3x - 4\):
\[ \lim_{x \to 2^+} f(x) = 3(2) - 4 = 2. \]
Since \(\text{LHL} \neq \text{RHL}\), the limit at \(x = 2\) does not exist.
If we were asked to calculate \(\lim_{x \to 2} f(x)\) for the piecewise function \(f(x)\) above,
simple substitution would give \(f(2) = 2\). However, the limits are:
\[ \lim_{x \to 2^-} f(x) = 3, \quad \lim_{x \to 2^+} f(x) = 2. \]Since \(\text{LHL} \neq \text{RHL}\), the limit \(\lim_{x \to 2} f(x)\) does not exist, and substitution would give an incorrect result.
-
-
Limit Point Not in the Domain:
If \(x = a\) is not part of the domain of \(f(x)\), the function cannot be evaluated directly at \(a\), and we must consider the behavior of \(f(x)\) as \(x\) approaches \(a\). e.g.
Consider the function \(f(x) = \ln x\). The domain of \(\ln x\) is \((0, \infty)\), so \(x = 0\) is not part of its domain. As \(x \to 0^+\), the natural logarithm becomes arbitrarily large in the negative direction because the logarithm of values close to \(0^+\) is a very large negative number. Thus,
\[ \lim_{x \to 0^+} \ln x = -\infty. \]In this case, direct substitution at \(x = 0\) is not possible because \(f(x)\) is undefined at \(x = 0\). Instead, we analyze the behavior of \(f(x)\) as \(x\) approaches \(0^+\).
Continuous functions make calculating limits simple because substitution works directly. However, when faced with indeterminate forms, infinity, discontinuities, or undefined points, we need to apply more advanced techniques to evaluate limits. These scenarios often lead us into deeper topics in calculus, which we will explore in detail.
Elementary Functions and Their Continuity
All functions we encounter in mathematics are typically constructed using elementary functions and basic operations like addition, subtraction, multiplication, division, and composition. Below is a summary of common elementary functions and their properties regarding continuity:
| Sr. No. | Function | Remarks on Continuity |
|---|---|---|
| 1 | Polynomial Functions | \(f(x) = a_0x^n + a_1x^{n-1} + \dots + a_n\), where \(a_0, a_1, \dots, a_n \in \mathbb{R}\) and \(n \in \mathbb{N} \cup \{0\}\). Continuous for all \(x \in \mathbb{R}\). |
| 2 | Exponential Functions | \(f(x) = a^x\), where \(a > 0\) and \(a \neq 1\). Continuous for all \(x \in \mathbb{R}\). |
| 3 | Logarithmic Functions | \(f(x) = \log_a x\), where \(a > 0, a \neq 1\). Continuous for all \(x \in (0, \infty)\). |
| 4 | Rational Functions | \(f(x) = \frac{p(x)}{q(x)}\), where \(p(x)\) and \(q(x)\) are polynomials. Continuous for all \(x \in \mathbb{R}\) except at points where \(q(x) = 0\). |
| 5 | Trigonometric Functions | - \(\sin x\) and \(\cos x\): Continuous for all \(x \in \mathbb{R}\). |
| - \(\tan x\) and \(\sec x\): Discontinuous at \(x = \frac{(2n+1)\pi}{2}, n \in \mathbb{Z}\). | ||
| - \(\cot x\) and \(\csc x\): Discontinuous at \(x = n\pi, n \in \mathbb{Z}\). | ||
| 6 | Inverse Trigonometric Functions | Functions like \(\arcsin x, \arccos x, \arctan x\), etc., are continuous within their respective domains. |
| 7 | Greatest Integer Function (\([x]\)) | Discontinuous at all \(x \in \mathbb{Z}\). |
| 8 | Fractional Part Function (\(\{x\} = x - [x]\)) | Discontinuous at all \(x \in \mathbb{Z}\). |
| 9 | Absolute Value Function | Continuous for all \(x \in \mathbb{R}\). |
Composition Rule for Continuous Functions
If \(f(x)\) is a continuous function and \(g(x)\) is a function for which \(\lim_{x \to a} g(x)\) exists, then:
This result holds because the continuity of \(f(x)\) ensures that the limit can "pass through" the function \(f\). Intuitively, \(f(x)\) does not disrupt the limiting behavior of \(g(x)\).
Example 1: \(\lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2\)
Here, \(f(x) = x^2\) is continuous, and \(g(x) = \frac{\sin x}{x}\) has the limit:
Using the composition rule:
Example 2: \(\lim_{x \to 0} e^{\frac{\sin x}{x}}\)
Here, \(f(x) = e^x\) is continuous, and \(g(x) = \frac{\sin x}{x}\) has the limit:
Using the composition rule:
Evaluating Limits of functions involving discontinuities
Before we delve into handling indeterminate forms, let us first practice evaluating limits for functions involving the greatest integer function (GIF), modulus, fractional part function, and piecewise-defined functions. These functions often require special attention due to their unique behaviors, particularly at critical points where their definitions change.
Example
Evaluate \(\lim_{x \to 1} \frac{[x] + 1}{1 + x}\), where \([x]\) denotes the greatest integer function. Does the limit exist?
Solution:
For the limit to exist at \(x = 1\), the left-hand limit (LHL) and the right-hand limit (RHL) must both exist and be equal. Since the greatest integer function \([x]\) changes its value at \(x = 1\), we cannot directly substitute \(x = 1\). Instead, we evaluate the LHL and RHL separately.
Step 1: Right-Hand Limit (\(x \to 1^+\))
As \(x \to 1^+\), \(x > 1\), which implies \(1 < x < 2\). In this interval, the greatest integer function satisfies \([x] = 1\), which is constant. The numerator of the given expression becomes:
The function simplifies to:
Now, as \(x \to 1^+\), the denominator \(1 + x\) approaches \(2\). The function is continuous in this region, so substituting \(x = 1\) directly gives:
Step 2: Left-Hand Limit (\(x \to 1^-\))
As \(x \to 1^-\), \(x < 1\), which implies \(0 < x < 1\). In this interval, the greatest integer function satisfies \([x] = 0\), which is constant. The numerator of the given expression becomes:
The function simplifies to:
Now, as \(x \to 1^-\), the denominator \(1 + x\) again approaches \(2\). The function is continuous in this region, so substituting \(x = 1\) directly gives:\
Step 3: Compare the LHL and RHL
The right-hand limit is:
The left-hand limit is:
Since the LHL and RHL are not equal, the limit does not exist at \(x = 1\).
Thus, the limit does not exist.
Rigorous Analysis for Pessimists
To evaluate \(\lim_{x \to 1} \frac{[x] + 1}{1 + x}\) with absolute rigor, we analyze the behavior of the function \(f(x) = \frac{[x] + 1}{1 + x}\) as \(x\) approaches \(1\) from both sides, using specific numerical sequences to ensure clarity and precision.
Right-Hand Behavior (\(x \to 1^+\)):
For \(x > 1\), the greatest integer function satisfies \([x] = 1\). Substituting into \(f(x)\), we have:
Testing with \(x = 1.1, 1.01, 1.001\):
Thus, as \(x \to 1^+\), \(f(x) \to 1\).
Left-Hand Behavior (\(x \to 1^-\)):
For \(x < 1\), the greatest integer function satisfies \([x] = 0\). Substituting into \(f(x)\), we have:
Testing with \(x = 0.9, 0.99, 0.999\):
Thus, as \(x \to 1^-\), \(f(x) \to 0.5\).
- The right-hand limit (\(\lim_{x \to 1^+} f(x)\)) is \(1\).
- The left-hand limit (\(\lim_{x \to 1^-} f(x)\)) is \(0.5\).
Since the left-hand limit and right-hand limit are not equal, the overall limit does not exist.
Example
Evaluate \(\lim_{x \to 0} \frac{[x] \sin x}{1 + \cos^2 x}\), where \([x]\) denotes the greatest integer function. Does the limit exist?
Solution:
To determine if the limit exists, we compute the left-hand limit (LHL) and the right-hand limit (RHL) separately since \([x]\) changes its value at \(x = 0\).
Right-Hand Limit (\(x \to 0^+\)):
As (\(x \to 0^+\)) , we have \(0 < x < 1\), so \([x] = 0\), which is constant. Substituting into the given expression:
Thus, the right-hand limit is:
Left-Hand Limit (\(x \to 0^-\)):
As (\(x \to 0^-\)), we have \(-1 < x < 0\), so \([x] = -1\), which is constant. Substituting into the given expression:
Now, as \(x \to 0^-\), \(\sin x \to 0\) and \(\cos^2 x \to 1\), so the denominator approaches \(1 + 1 = 2\). Hence:
The left-hand limit and right-hand limit are equal:
Thus, the overall limit exists, and:
Example
Evaluate \(\lim_{x \to 2} [x^2 - 2x]\), where \([x]\) denotes the greatest integer function. Does the limit exist?
Solution:
The expression involves the greatest integer function \([x]\), which takes the largest integer less than or equal to \(x^2 - 2x\). At \(x = 2\), the term \(x^2 - 2x = x(x - 2)\) evaluates exactly to \(0\), an integer. This indicates that, to determine the behavior of \([x^2 - 2x]\) as \(x \to 2\), we must separately analyze the left-hand limit (LHL) and right-hand limit (RHL).
For \(x \to 2^+\), we observe that \(x > 2\). In this case, \(x - 2 > 0\), so \(x^2 - 2x = x(x - 2)\) becomes a small positive value approaching \(0^+\) as \(x \to 2^+\). The greatest integer function \([x^2 - 2x]\) therefore evaluates to \(0\), as \(0^+\) rounds down to \(0\) in the greatest integer definition. Thus,
For \(x \to 2^-\), we have \(x < 2\), so \(x - 2 < 0\). In this case, \(x^2 - 2x = x(x - 2)\) becomes a small negative value approaching \(0^-\) as \(x \to 2^-\). The greatest integer function \([x^2 - 2x]\) evaluates to \(-1\), as \(0^-\) rounds down to \(-1\). Thus,
[
\lim_{x \to 2^-} [x^2 - 2x] = \lim_{x \to 2^-} -1 = -1.
]
Since the left-hand limit (\(-1\)) is not equal to the right-hand limit (\(0\)), the two-sided limit does not exist. The discontinuity of the greatest integer function at non-integer values leads to this result.
Thus, the limit \(\lim_{x \to 2} [x^2 - 2x]\) does not exist because the left-hand and right-hand limits are not equal.
Below is the graph of \(y = x^2-2x\):
Example
Evaluate \(\lim_{x \to 5/2} [x + 1/2] - [x - 3/2]\), where \([x]\) denotes the greatest integer function. Does the limit exist?
Solution:
The expression involves the greatest integer function \([x]\), which takes the largest integer less than or equal to its argument. To determine whether the limit exists, we analyze the left-hand limit (LHL) and right-hand limit (RHL) separately.
For \(x \to 5/2^+\):
Similarly,
Substituting into the expression, we get:
Thus, \(\lim_{x \to 5/2^+} [x + 1/2] - [x - 3/2] = 2\).
For \(x \to 5/2^-\):
Similarly,
Substituting into the expression, we get:
Thus, \(\lim_{x \to 5/2^-} [x + 1/2] - [x - 3/2] = 2\).
Since the left-hand limit and right-hand limit are equal, the limit exists, and:
Example
Evaluate \(\lim_{x \to \pi/6^+} [\tan(x + \pi/6)]^{\tan(3x)}\), where \([x]\) denotes the greatest integer function.
Solution:
To evaluate the limit, we carefully analyze the behavior of the base \([\tan(x + \pi/6)]\) and the exponent \(\tan(3x)\) as \(x \to \pi/6^+\).
As \(x \to \pi/6^+\), the term \(x + \pi/6 \to \pi/3^+\). Since \(\tan(x)\) is continuous at \(\pi/3\), it follows that \(\tan(x + \pi/6) \to \tan(\pi/3) = \sqrt{3}\). Applying the greatest integer function to this value, we find:
Thus, as \(x \to \pi/6^+\), the base of the given expression remains constant and equal to \(1\).
Next, we analyze the exponent \(\tan(3x)\). As \(x \to \pi/6^+\), we have \(3x \to \pi/2^+\). Near \(x = \pi/2^+\), \(\tan(x)\) diverges to \(-\infty\). Therefore, as \(x \to \pi/6^+\),
Substituting these results into the given expression, we write:
Since the base is \(1\), and \(1^u = 1\) for any real number \(u\), the value of the limit simplifies to:
Conclusion:
The limit exists and evaluates to:
To make this reasoning clearer, observe that as \(x \to \pi/6^+\), through any sequence approaching \(\pi/6^+\), the term \(\tan(x + \pi/6)\) always remains very close to \(\sqrt{3}\). Since the greatest integer function \([x]\) takes the largest integer less than or equal to its argument, we find that \([\tan(x + \pi/6)] = 1\) for all \(x\) sufficiently close to \(\pi/6^+\).
It is important to note that the base of the expression, \([\tan(x + \pi/6)]\), is constant and equal to \(1\); it is not merely "approaching" \(1\). The constancy of the base ensures that any power of it, no matter how large, small, or divergent the exponent may be, always evaluates to \(1\).
Thus, near \(x = \pi/6\), the entire expression simplifies to a constant value \(1\). In other words, for all \(x\) sufficiently close to \(\pi/6^+\), the given expression is effectively equal to \(1\). This constant behavior of the limit is the reason the expression evaluates to:
Using Special Substitutions to evaluate One-sided Limits
Evaluating Limits Using Substitutions
Sometimes, evaluating one-sided limits becomes easier through appropriate substitutions. Consider the right-hand limit \(\lim_{x \to a^+} f(x)\). Here, \(x \to a^+\) implies that \(x > a\). Let the distance between \(a\) and \(x\) be a positive number \(h > 0\). Then:
As \(x \to a^+\), it follows that \(h \to 0^+\). Substituting \(x = a + h\), the limit can be rewritten as:
This substitution often simplifies the given expression and makes the evaluation of the limit more straightforward.
Similarly, for the left-hand limit \(\lim_{x \to a^-} f(x)\), we note that \(x < a\). Let the distance between \(a\) and \(x\) be a positive number \(h > 0\). Then:
As \(x \to a^-\), it follows that \(h \to 0^+\). Substituting \(x = a - h\), the limit can be rewritten as:
These substitutions, \(x = a + h\) for \(x \to a^+\) and \(x = a - h\) for \(x \to a^-\), are particularly useful for simplifying expressions involving functions that depend on \(x\) relative to a fixed point \(a\). By working with \(h\), the distance from \(a\), many limits become easier to analyze and compute.
Example
Evaluate \(\lim_{x \to 3\pi/4} [\sin x + \cos x]\), where \([x]\) denotes the greatest integer function.
Solution:
To determine the behavior of the limit, we compute the right-hand limit (RHL) and the left-hand limit (LHL) separately.
Right-Hand Limit (\(x \to 3\pi/4^+\)):
Substituting \(x = 3\pi/4 + h\) into the given expression:
Using the angle addition formulas:
Substituting \(\sin(3\pi/4) = \frac{1}{\sqrt{2}}\) and \(\cos(3\pi/4) = -\frac{1}{\sqrt{2}}\):
Simplifying:
Thus, the limit becomes:
As \(h \to 0^+\), \(\sin h \to 0^+\), so \(-\sqrt{2}\sin h \to 0^-\). Therefore:
Left-Hand Limit (\(x \to 3\pi/4^-\)):
Substituting \(x = 3\pi/4 - h\) into the given expression:
Using the angle subtraction formulas:
Substituting \(\sin(3\pi/4) = \frac{1}{\sqrt{2}}\) and \(\cos(3\pi/4) = -\frac{1}{\sqrt{2}}\):
Simplifying:
Thus, the limit becomes:
As \(h \to 0^+\), \(\sin h \to 0^+\), so \(\sqrt{2}\sin h \to 0^+\). Therefore:
Conclusion:
The right-hand limit is \(-1\), and the left-hand limit is \(0\). Since the two limits are not equal, the overall limit does not exist: