Evaluating ∞/∞ Form

Introduction

Consider the limit

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

where both \( f(x) \to \pm\infty \) and \( g(x) \to \pm\infty \) as \( x \to a \), resulting in an indeterminate form of type \( \frac{\infty}{\infty} \).

To analyze this, we rewrite the given limit in an equivalent form by considering its reciprocal transformation. Define

\[ h(x) = \frac{1}{g(x)}. \]

Since \( g(x) \to \pm\infty \), it follows that \( h(x) \to 0 \). Similarly, considering

\[ k(x) = \frac{1}{f(x)}, \]

we observe that since \( f(x) \to \pm\infty \), it follows that \( k(x) \to 0 \). This transforms the original limit into

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{\frac{1}{g(x)}}{\frac{1}{f(x)}} = \lim_{x \to a} \frac{h(x)}{k(x)}. \]

Since both \( h(x) \to 0 \) and \( k(x) \to 0 \), we now obtain an indeterminate form of type \( \frac{0}{0} \).

The Indeterminate Nature this form

The indeterminate nature of the \( \frac{\infty}{\infty} \) form arises because knowing that both functions \( f(x) \) and \( g(x) \) tend to infinity does not, by itself, determine the behavior of their ratio. The essential question is how fast each function approaches infinity relative to the other.

Consider the limit

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

where both \( f(x) \to \infty \) and \( g(x) \to \infty \). The result depends on the relative growth rates of \( f(x) \) and \( g(x) \), leading to different possible outcomes:

1. A finite nonzero limit: If \( f(x) \) and \( g(x) \) grow at comparable rates, their ratio may tend to a finite number.

   • Example: \( \lim_{x \to \infty} \frac{5x}{2x} = \frac{5}{2} \), since both functions are linear and have the same degree.

2. The limit is \( \infty \) or \( -\infty \): If \( f(x) \) grows much faster than \( g(x) \), their ratio tends to infinity.

   • Example: \( \lim_{x \to \infty} \frac{x^3}{x} = \infty \), since \( x^3 \) dominates \( x \) as \( x \to \infty \).

3. The limit is 0: If \( g(x) \) grows much faster than \( f(x) \), their ratio tends to zero.

   • Example: \( \lim_{x \to \infty} \frac{x}{x^3} = 0 \), since \( x^3 \) dominates \( x \).

These different possibilities show why \( \frac{\infty}{\infty} \) is an indeterminate form—it does not uniquely determine a limit. The fraction's behavior depends on the comparative growth rates of \( f(x) \) and \( g(x) \), requiring further analysis.

Resolving \( \frac{\infty}{\infty} \) Forms by Reciprocal Transformation

One way to handle limits of the form \( \frac{\infty}{\infty} \) is to convert them into an equivalent \( \frac{0}{0} \) form by taking reciprocals. However, this approach is rarely used in practice, as more direct simplifications are often available. Nonetheless, in some cases, recognizing this transformation can lead to an alternative method of evaluating limits.

Consider the limit

\[ \lim_{x \to 0} \frac{\cot 2x}{\cot x}. \]

Since \( \cot x = \frac{\cos x}{\sin x} \), we observe that as \( x \to 0 \), both \( \cot 2x \to \infty \) and \( \cot x \to \infty \), resulting in an \( \frac{\infty}{\infty} \) indeterminate form.

Taking reciprocals,

\[ \lim_{x \to 0} \frac{\cot 2x}{\cot x} = \lim_{x \to 0} \frac{1/\tan 2x}{1/\tan x} = \lim_{x \to 0} \frac{\tan x}{\tan 2x}. \]

Now, using the standard limit result,

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

the expression can be analyzed further.

Resolving \( \frac{\infty}{\infty} \) Using L'Hôpital’s Rule

We have already introduced L’Hôpital’s Rule and its application in resolving indeterminate forms. This method is particularly useful and frequently employed whenever we encounter the \( \frac{\infty}{\infty} \) form. The core idea is to differentiate the numerator and denominator separately and then evaluate the resulting limit. This process can be repeated as needed, provided the conditions for L'Hôpital’s Rule hold at each step.

Consider the limit

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

Both the numerator and denominator tend to \( \infty \), leading to the \( \frac{\infty}{\infty} \) form. Applying L'Hôpital’s Rule, we differentiate the numerator and denominator separately:

\[ \frac{d}{dx} (x^2) = 2x, \quad \frac{d}{dx} (2x^2 - x + 1) = 4x - 1. \]

Thus, the limit transforms into

\[ \lim_{x \to \infty} \frac{2x}{4x - 1}. \]

Since this still results in the \( \frac{\infty}{\infty} \) form, we apply L'Hôpital’s Rule again:

\[ \frac{d}{dx} (2x) = 2, \quad \frac{d}{dx} (4x - 1) = 4. \]

This reduces the limit to

\[ \lim_{x \to \infty} \frac{2}{4} = \frac{1}{2}. \]

Thus, by applying L'Hôpital’s Rule twice, we conclude that

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

This example illustrates how L'Hôpital’s Rule simplifies indeterminate forms step by step, effectively reducing complexity and allowing us to compute the limit efficiently.

Resolving \( \frac{\infty}{\infty} \) Using Dominant Term Analysis

Consider the limit

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

where both \( f(x) \to \infty \) and \( g(x) \to \infty \) as \( x \to a \), resulting in an indeterminate form of type \( \frac{\infty}{\infty} \). The behavior of this limit depends on the relative growth rates of \( f(x) \) and \( g(x) \). That is, the limit is determined by how fast \( f(x) \) and \( g(x) \) approach infinity with respect to each other.

If \( f(x) \) and \( g(x) \) grow at comparable rates, their ratio may tend to a finite constant \( c \), where

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = c, \quad c \neq 0, \pm\infty. \]

If \( f(x) \) grows significantly faster than \( g(x) \), then

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

In this case, we say that \( f(x) \) is the dominant term, meaning that \( f(x) \) increases unboundedly at a higher rate than \( g(x) \).

Conversely, if \( g(x) \) grows significantly faster than \( f(x) \), then

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

Here, \( g(x) \) is dominant, forcing the fraction to approach zero.

Polynomial vs Exponential Growth

Consider the limit

\[ \lim_{x \to \infty} \frac{x^\alpha}{a^x}, \quad a > 1, \quad \alpha > 0. \]

Since both the numerator \( x^\alpha \) and the denominator \( a^x \) tend to infinity as \( x \to \infty \), we encounter an indeterminate form of type \( \frac{\infty}{\infty} \). Applying L'Hôpital’s Rule, we differentiate the numerator and denominator:

\[ \lim_{x \to \infty} \frac{\alpha x^{\alpha - 1}}{a^x \ln a}. \]

If \( \alpha - 1 \) remains positive, the limit continues to exhibit the \( \frac{\infty}{\infty} \) form, warranting repeated applications of L'Hôpital’s Rule. Upon successive differentiation, after \( k \) iterations, we obtain:

\[ \lim_{x \to \infty} \frac{\alpha (\alpha - 1) (\alpha - 2) \cdots (\alpha - k + 1) x^{\alpha - k}}{a^x (\ln a)^k}. \]

At this stage, when \( \alpha - k < 0 \), the power of \( x \) in the numerator becomes negative, rendering \( x^{\alpha - k} \) as a decreasing function towards zero as \( x \to \infty \). Meanwhile, the exponential term \( a^x \) in the denominator continues its unbounded growth, now amplified by the factor \( (\ln a)^k \).

Thus, the numerator approaches a constant value while the denominator increases exponentially. The overwhelming growth of the denominator forces the limit to converge to zero:

\[ \lim_{x \to \infty} \frac{x^\alpha}{a^x} = 0. \]

This analysis demonstrates that exponential functions \( a^x \) (where \( a > 1 \)) grow faster than any polynomial \( x^\alpha \) as \( x \to \infty \). The dominance of exponential growth over polynomial growth is evident, leading us to the swift conclusion such as:

\[ \lim_{x \to \infty} \frac{x^{100000}}{(1.00001)^x} = 0. \]

This principle underscores the inherent strength of exponential growth, illustrating that regardless of the degree of the polynomial, the exponential function will eventually dominate as \( x \) becomes large.

Logarithmic vs Polynomial

Consider the limit

\[ \lim_{x \to \infty} \frac{\ln x}{x^\alpha}, \quad \alpha > 0. \]

Since both \( \ln x \to \infty \) and \( x^\alpha \to \infty \) as \( x \to \infty \), the given expression is of the indeterminate form \( \frac{\infty}{\infty} \), allowing the application of L'Hôpital’s Rule. Differentiating the numerator and denominator separately, we obtain

\[ \lim_{x \to \infty} \frac{1/x}{\alpha x^{\alpha - 1}}. \]

Rewriting the fraction,

\[ \lim_{x \to \infty} \frac{1}{\alpha x^\alpha}. \]

Since \( x^\alpha \to \infty \) for \( \alpha > 0 \), the denominator grows unbounded, forcing the fraction to approach zero:

\[ \lim_{x \to \infty} \frac{\ln x}{x^\alpha} = 0. \]

Thus, polynomial growth dominates logarithmic growth as \( x \to \infty \), demonstrating that \( x^\alpha \) increases significantly faster than \( \ln x \).

Since we have established that \( \ln x \) grows significantly slower than any polynomial \( x^\alpha \) for \( \alpha > 0 \), we can conclude without hesitation that

\[ \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = 0. \]

Here, \( \sqrt{x} = x^{1/2} \) is a polynomial function with exponent \( \alpha = 1/2 \), which still satisfies the previously proven result. The polynomial term \( \sqrt{x} \) grows unbounded, while \( \ln x \) remains asymptotically negligible in comparison, forcing the fraction to vanish as \( x \to \infty \).

From the results established above, we can conclude that as \( x \to \infty \), the logarithmic function grows significantly slower than any polynomial, which in turn grows slower than any exponential function. This can be expressed as

\[ \ln x \ll x^\alpha \ll a^x, \quad \text{for } \alpha > 0, \quad a > 1 \text{ and large } x. \]

Resolving \( \frac{\infty}{\infty} \) Using the Dominant Term

Consider the limit

\[ \lim_{x \to a} \frac{a_1 f_1(x) + a_2 f_2(x)}{b_1 g_1(x) + b_2 g_2(x)}. \]

Since both the numerator and denominator tend to infinity, the expression takes the indeterminate form \( \frac{\infty}{\infty} \). This implies that at least one function in the numerator and denominator is responsible for the divergence. If we can determine which function grows the fastest as \( x \to a \), we can factor it out from both numerator and denominator to simplify the expression and resolve the limit.

To illustrate this, consider the specific example

\[ \lim_{x \to \infty} \frac{2^x + 5 \cdot 3^x}{3^x + 7x}. \]

Since exponential functions grow faster than polynomials, and among \( 2^x \) and \( 3^x \), the term \( 3^x \) dominates because its base is larger, we divide both numerator and denominator by \( 3^x \):

\[ \lim_{x \to \infty} \frac{\frac{2^x}{3^x} + 5}{1 + \frac{7x}{3^x}}. \]

Rewriting \( 2^x/3^x = (2/3)^x \), we obtain:

\[ \lim_{x \to \infty} \frac{(2/3)^x + 5}{1 + 7x/3^x}. \]

Since \( (2/3)^x \to 0 \) and \( 7x/3^x \to 0 \) as \( x \to \infty \), the expression simplifies to

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

Thus,

\[ \lim_{x \to \infty} \frac{2^x + 5 \cdot 3^x}{3^x + 7x} = 5. \]

This approach of identifying the dominant term and normalizing by it provides an efficient way to resolve \( \frac{\infty}{\infty} \) forms without differentiation.

Behavior of Rational Functions as \( x \to \infty \)

Consider the limit at infinity of a general rational function

\[ \lim_{x \to \infty} \frac{a_0 x^m + a_1 x^{m-1} + \dots + a_m}{b_0 x^n + b_1 x^{n-1} + \dots + b_n}, \]

where \( a_0 \neq 0 \) and \( b_0 \neq 0 \). To analyze the limit, we factor out \( x^m \) from the numerator and \( x^n \) from the denominator. In the numerator, this yields

\[ a_0 x^m \Bigl(1 + \frac{a_1}{a_0} \frac{1}{x} + \dots + \frac{a_m}{a_0} \frac{1}{x^m}\Bigr), \]

and in the denominator,

\[ b_0 x^n \Bigl(1 + \frac{b_1}{b_0} \frac{1}{x} + \dots + \frac{b_n}{b_0} \frac{1}{x^n}\Bigr). \]

Thus, the limit becomes

\[ \lim_{x \to \infty} \frac{a_0 x^m \bigl(1 + \frac{a_1}{a_0}\frac{1}{x} + \dots + \frac{a_m}{a_0}\frac{1}{x^m}\bigr)}{b_0 x^n \bigl(1 + \frac{b_1}{b_0}\frac{1}{x} + \dots + \frac{b_n}{b_0}\frac{1}{x^n}\bigr)} = \lim_{x \to \infty} \frac{a_0}{b_0} \, x^{m-n} \, \frac{1 + \frac{a_1}{a_0}\frac{1}{x} + \dots + \frac{a_m}{a_0}\frac{1}{x^m}}{1 + \frac{b_1}{b_0}\frac{1}{x} + \dots + \frac{b_n}{b_0}\frac{1}{x^n}}. \]

As \( x \to \infty \), the factors \(\frac{1}{x}, \frac{1}{x^2}, \dots\) all approach 0, so the fractional part in the above expression tends to 1. Consequently, the behavior of the limit is governed by \( x^{m-n} \) and the constant factor \( \frac{a_0}{b_0} \):

1. If \( m = n \), then \( x^{m-n} = x^0 = 1 \). Hence,

   \[ \lim_{x \to \infty} \frac{a_0 x^m + \dots}{b_0 x^n + \dots} = \frac{a_0}{b_0}. \]

2. If \( m < n \), then \( x^{m-n} \to 0 \). Since the remaining fraction tends to 1,

   \[ \lim_{x \to \infty} \frac{a_0 x^m + \dots}{b_0 x^n + \dots} = 0. \]

3. If \( m > n \), then \( x^{m-n} \to \infty \). The sign of the limit depends on the sign of \(\frac{a_0}{b_0}\):

   • If \(\frac{a_0}{b_0} > 0\), the limit is \(+\infty\).
   • If \(\frac{a_0}{b_0} < 0\), the limit is \(-\infty\).

Hence, the leading powers of \( x \) in the numerator and denominator, along with their leading coefficients, determine the behavior of rational functions as \( x \) tends to infinity.

Example

Consider the limit

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

The numerator is a cubic polynomial with highest degree term \(x^3\) and leading coefficient 1, while the denominator has highest degree term \(-3x^3\) and leading coefficient \(-3\). When a rational function’s numerator and denominator share the same highest degree, the limit as \(x \to \infty\) is the ratio of the leading coefficients. Here, that ratio is

\[ \frac{1}{-3} = -\frac{1}{3}. \]

Hence,

\[ \lim_{x \to \infty} \frac{(x - 1)(x - 2)(x - 3)}{1 - 3x^3} = -\frac{1}{3}. \]

Example

Evaluate

\[ \lim_{x \to -\infty} \frac{(x + 1)(x - 1)(x - 2)}{(1 - 2x)^2 (x + 3)}. \]

Solution.

Set \(x = -y\), so that \(y \to \infty\) as \(x \to -\infty\). Inside the limit operator,

\[ \lim_{x \to -\infty} \frac{(x + 1)(x - 1)(x - 2)}{(1 - 2x)^2 (x + 3)} = \lim_{y \to \infty} \frac{\bigl((-y) + 1\bigr)\bigl((-y) - 1\bigr)\bigl((-y) - 2\bigr)} {\bigl(1 - 2(-y)\bigr)^2 \bigl((-y) + 3\bigr)}. \]

The numerator factors as

\[ (-y + 1)(-y - 1)(-y - 2) = (-1)^3 (y - 1)(y + 1)(y + 2) = -\,\bigl[(y - 1)(y + 1)(y + 2)\bigr]. \]

The denominator becomes

\[ (1 + 2y)^2 \bigl(-y + 3\bigr) = -(1 + 2y)^2 (y - 3). \]

Hence,

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

For large \(y\), the highest-order term in \((y - 1)(y + 1)(y + 2)\) is \(y^3\), and in \((1 + 2y)^2 (y - 3)\) it is \(4y^3\). Thus,

\[ \lim_{y \to \infty} \frac{(y - 1)(y + 1)(y + 2)}{(1 + 2y)^2 (y - 3)} = \lim_{y \to \infty} \frac{y^3 + \dots}{4y^3 + \dots} = \frac{1}{4}. \]

Therefore,

\[ \lim_{x \to -\infty} \frac{(x + 1)(x - 1)(x - 2)}{(1 - 2x)^2 (x + 3)} = \frac{1}{4}. \]