Evaluating one raised to power infinity form

Introduction

Consider the limit

\[ \lim_{x \to a} \bigl(f(x)\bigr)^{g(x)} \]

under the assumptions that \(\lim_{x \to a} f(x) = 1\) and \(\lim_{x \to a} g(x) = \pm\infty\). This creates the indeterminate form \(1^\infty\), meaning that the limiting behavior is uncertain and depends on the precise relationship between \(f(x)\) and \(g(x)\).

To analyze this form rigorously, we take the natural logarithm and rewrite the expression using exponentiation. By the identity

\[ a^b = e^{b \ln a}, \]

we transform the given limit into

\[ \lim_{x \to a} \bigl(f(x)\bigr)^{g(x)} = \lim_{x \to a} e^{g(x) \ln f(x)}. \]

The key observation is that since \(f(x) \to 1\), its logarithm satisfies \(\ln f(x) \to 0\). Meanwhile, \(g(x) \to \pm\infty\), so the exponent \(g(x) \ln f(x)\) takes the form \(0 \times \infty\), which we already recognize as an indeterminate form. This confirms that \(1^\infty\) is truly indeterminate—it can lead to different outcomes depending on how quickly \(f(x)\) approaches 1 and how \(g(x)\) behaves.

Intuitive understanding of one raised to power infinity form

The indeterminate form \(1^\infty\) arises when we evaluate limits of the type

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

where \( f(x) \to 1 \) and \( g(x) \to \pm\infty \). The uncertainty comes from the conflicting tendencies of the base and the exponent.

Case 1: \( f(x) \to 1^+ \) and \( g(x) \to +\infty \)

Here, \( f(x) \) is slightly greater than 1 and is trying to push the expression towards 1. However, the exponent \( g(x) \) is growing larger and larger, pulling the value towards \(\infty\), since any number slightly greater than 1, when raised to an extremely large power, tends to infinity. The final outcome depends on the relative rates at which \( f(x) \) approaches 1 and \( g(x) \) approaches infinity. If \( f(x) \) approaches 1 very slowly while \( g(x) \) grows extremely fast, the result will be a large number, possibly even infinity. If \( f(x) \) approaches 1 fast enough, the result can be a finite number greater than 1.

Thus, in this case, the limit can take any value in the range \([1, \infty)\).

Case 2: \( f(x) \to 1^- \) and \( g(x) \to +\infty \)

Now, \( f(x) \) is slightly less than 1, so it is trying to push the expression towards 1. However, when a base less than 1 is raised to a large positive power, the exponent pulls the value towards 0. Again, the final result depends on how fast \( f(x) \) approaches 1 compared to how fast \( g(x) \) approaches infinity. If \( f(x) \) is approaching 1 very slowly, the result will be close to 0. If \( f(x) \) is approaching 1 very quickly, the result might still be close to 1.

Thus, in this case, the limit can take any value in the range \([0,1]\).

The indeterminate nature of \(1^\infty\) arises because two opposing forces are at play: the base trying to bring the expression to 1 and the exponent trying to pull it away. The final value depends on the relative speeds at which \( f(x) \to 1 \) and \( g(x) \to \infty \), leading to a wide range of possible outcomes.

How to evaluate one raised to power infinity form

To resolve the limit, we now evaluate

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

If this limit exists and evaluates to \(L\), then

\[ \lim_{x \to a} e^{g(x) \ln f(x)} = e^L. \]

Example

Evaluate the limit

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

Solution:

As \(x \to 0\), the base \(1 + x\) approaches \(1\), while the exponent \(1/x\) tends to \(+\infty\) if \(x \to 0^+\) and \(-\infty\) if \(x \to 0^-\). Thus, the expression is of the form \(1^\infty\), an indeterminate form.

Applying base change property of exponentials,

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

Since, \(e^x\) is a constinuous function, we can write,

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

Using the standard limit result

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

we obtain

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

Thus, the given limit evaluates to

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

We seldom evaluate \(1^\infty\) limits directly in the above example. Instead, we use a simplification that makes the process more systematic:

\[ \lim_{x \to a} f(x)^{g(x)} = \lim_{x \to a} e^{g(x) \ln f(x)} \]

where \(f(x) \to 1\) and \(g(x) \to \pm\infty\). To simplify further, we define \(t = f(x) - 1\), which approaches \(0\) as \(x \to a\). This substitution allows us to rewrite \(\ln f(x)\) in terms of \(t\):

\[ \lim_{x \to a} e^{g(x) \ln f(x)} = \lim_{x \to a} e^{g(x) \ln(1 + t)}. \]

Now, we use the standard result:

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

To apply this, we add and subtract 1, multiply and divide by \(t\) to complete the standard form:

\[ = \lim_{x \to a} e^{g(x) \cdot \frac{\ln(1 + t)}{t} \cdot t}. \]

Since \(\frac{\ln(1+t)}{t} \to 1\) as \(t \to 0\), we substitute:

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

Thus, we arrive at the key simplification:

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

If \(\lim_{x \to a} f(x) = 1\) and \(\lim_{x \to a} g(x) = \pm\infty\), then

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

This transformation converts the \(1^\infty\) form into the simpler indeterminate form \(0 \times \infty\), which can be analyzed using standard limit techniques.

Example

Evaluate the limit

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

Solution. Since \( \lim_{x \to \infty} \left(1 + \frac{1}{x}\right) = 1 \) and \( \lim_{x \to \infty} x = \infty \), this limit is of the indeterminate form \(1^\infty\). Using the theorem,

\[ \lim_{x \to \infty} f(x)^{g(x)} = \lim_{x \to \infty} e^{g(x) (f(x) - 1)}, \]

where \( f(x) = 1 + \frac{1}{x} \) and \( g(x) = x \), we rewrite:

\[ \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = \lim_{x \to \infty} e^{x \cdot \left(1 + \frac{1}{x} - 1\right)}. \]

This simplifies to

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

Thus,

\[ \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e. \]

Example

Evaluate the limit

\[ \lim_{x \to \frac{\pi}{2}} \bigl(1 + \cos x\bigr)^{\tan x}. \]

Solution. Since \(1 + \cos x \to 1\) and \(\tan x \to \infty\) as \(x \to \frac{\pi}{2}\), the given expression is of the indeterminate form \(1^\infty\). Using the standard theorem,

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

where \( f(x) = 1 + \cos x \) and \( g(x) = \tan x \), we rewrite:

\[ \begin{align} \lim_{x \to \frac{\pi}{2}} (1 + \cos x)^{\tan x} &= \lim_{x \to \frac{\pi}{2}} e^{\tan x \cdot (1+\cos x -1)} \\ &= \lim_{x \to \frac{\pi}{2}} e^{\tan x \cdot (\cos x)} \\ &= \lim_{x \to \frac{\pi}{2}} e^{\frac{\sin x}{\cos x} \cdot \cos x} \\ &= \lim_{x \to \frac{\pi}{2}} e^{\sin x}. \end{align} \]

Since \( \sin x \to 1 \) as \( x \to \frac{\pi}{2} \), we obtain:

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

Thus,

\[ \lim_{x \to \frac{\pi}{2}} \bigl(1 + \cos x\bigr)^{\tan x} = e. \]

Example

Evaluate the limit

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

Solution. Since

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

and the exponent \( x \to \infty \), the given expression is of the indeterminate form \(1^\infty\). Using the standard theorem,

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

where \( f(x) = \frac{x}{x+2} \) and \( g(x) = x \), we proceed as follows:

\[ \begin{align} \lim_{x \to \infty} \left(\frac{x}{x+2}\right)^x &= \lim_{x \to \infty} e^{x \cdot \left(\frac{x}{x+2} - 1\right)} \\ &= \lim_{x \to \infty} e^{x \cdot \left(\frac{x - (x+2)}{x+2}\right)} \\ &= \lim_{x \to \infty} e^{x \cdot \left(\frac{x - x -2}{x+2}\right)} \\ &= \lim_{x \to \infty} e^{x \cdot \left(\frac{-2}{x+2}\right)} \\ &= \lim_{x \to \infty} e^{- \frac{2x}{x+2}}. \end{align} \]

Since \(\lim_{x \to \infty} \frac{2x}{x+2} = 2\), we obtain:

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

Thus,

\[ \lim_{x \to \infty} \left(\frac{x}{x+2}\right)^x = e^{-2}. \]