Recovering Power Limits from Logarithmic Limits

From a power to a logarithmic limit

A variable power can be difficult to analyze directly because both its base and its exponent may change as $x$ approaches a point. For a positive base, logarithms give a reliable way to separate those two parts. The key is to find the limit of the exponent multiplied by the logarithm of the base, and then exponentiate that limit.

This tutorial concerns the recovery step: once the logarithmic expression has a limit, what does that tell us about the original power? The logarithm does not need to be defined at the limiting point itself. It is enough that the base is positive for $x$ sufficiently close to the point, on the part of the domain under consideration.

Definition: If $f(x)>0$, a real variable power $f(x)^{g(x)}$ is defined by $$ f(x)^{g(x)}=\exp\bigl(g(x)\ln f(x)\bigr). $$ Here $\ln$ is the natural logarithm and $\exp$ is the exponential function.

Thus, to study $f(x)^{g(x)}$, define the logarithmic expression $q(x)=g(x)\ln f(x)$. The original power is exactly $\exp(q(x))$. The crucial condition is positivity of the base: for general real exponents, this definition and the real logarithm require $f(x)>0$.

The recovery theorem

Theorem: Suppose $f(x)>0$ near the point being approached, and let $q(x)=g(x)\ln f(x)$. If $$ \lim_{x\to a}q(x)=L $$ for a finite real number $L$, then $$ \lim_{x\to a}f(x)^{g(x)}=e^L. $$ The same conclusion holds for one-sided limits and limits as $x\to\infty$ or $x\to-\infty$, provided the stated logarithmic limit exists.

Justification. For every $x$ in the relevant domain, the definition of the real power gives $f(x)^{g(x)}=\exp(q(x))$. The exponential function is continuous at every real number, so when $q(x)\to L$, continuity gives $\exp(q(x))\to\exp(L)=e^L$. Substituting the identity for the power proves the result.

This is a composition argument: the inner expression $q(x)$ approaches $L$, and the continuous outer function $\exp$ carries that limit to $e^L$. No separate limit law for powers is needed.

The same reasoning also describes infinite logarithmic limits. If $q(x)\to+\infty$, then $\exp(q(x))\to+\infty$; if $q(x)\to-\infty$, then $\exp(q(x))\to0$. These follow from the behavior of the exponential function at the two ends of the real line. Consequently, the logarithmic limit can determine whether the power approaches a positive finite value, grows without bound, or tends to zero.

Worked examples

Worked Example: A finite limit obtained using L'Hôpital's rule

Find $\displaystyle\lim_{x\to\infty}(x^2+3x)^{1/\ln x}$.

For $x$ sufficiently large, $x^2+3x>0$ and $\ln x>0$, so the power is defined as a real number. Its logarithmic expression is

q(x)=\frac{\ln(x^2+3x)}{\ln x}.

As $x\to\infty$, the numerator and denominator both tend to $+\infty$, so L'Hôpital's rule applies. Differentiating the numerator and denominator gives

\lim_{x\to\infty}q(x) = \lim_{x\to\infty} \frac{(2x+3)/(x^2+3x)}{1/x} = \lim_{x\to\infty}\frac{x(2x+3)}{x^2+3x}.

Expanding the numerator and dividing both numerator and denominator by $x^2$,

\frac{x(2x+3)}{x^2+3x} = \frac{2x^2+3x}{x^2+3x} = \frac{2+3/x}{1+3/x} \longrightarrow \frac{2+0}{1+0}=2.

The recovery theorem now gives

\lim_{x\to\infty}(x^2+3x)^{1/\ln x}=e^2.

Check: The logarithm of the original power is exactly $q(x)$, whose limit is $2$. Exponentiating this limiting logarithm gives $e^2$, as claimed.

Worked Example: A power tending to one

Find $\displaystyle\lim_{x\to\infty}(5+1/x)^{1/x}$.

For $x>0$, the base is positive. Take the logarithm of the power by multiplying the exponent by the logarithm of the base:

q(x)=\frac{\ln(5+1/x)}{x}.

As $x\to\infty$, $5+1/x\to5$, so continuity of the logarithm at the positive number $5$ gives $\ln(5+1/x)\to\ln5$, a finite value. Meanwhile $x\to+\infty$. Therefore

q(x)=\frac{\ln(5+1/x)}{x}\longrightarrow 0.

Applying the theorem,

\lim_{x\to\infty}(5+1/x)^{1/x}=e^0=1.

Check: The logarithm of this power tends to $0$, and the exponential of $0$ is $1$. The base itself tends to $5$, but the power's limit is determined by the product of the exponent and the logarithm of the base, not by the base alone.

Worked Example: A power tending to zero

Find $\displaystyle\lim_{x\to0^+}x^{1/\sqrt{\ln(1/x)}}$.

For $0<x<1$, $\ln(1/x)>0$, so the square root is real and the base $x$ is positive. The logarithmic expression is

q(x)=\frac{\ln x}{\sqrt{\ln(1/x)}}.

Since $\ln x=-\ln(1/x)$, this simplifies to

q(x) = -\frac{\ln(1/x)}{\sqrt{\ln(1/x)}} = -\sqrt{\ln(1/x)}.

As $x\to0^+$, $1/x\to+\infty$, so $\ln(1/x)\to+\infty$ and hence $q(x)\to-\infty$. The exponential of a quantity tending to $-\infty$ tends to zero. Therefore

\lim_{x\to0^+}x^{1/\sqrt{\ln(1/x)}}=0.

Check: The logarithm of the positive power tends to $-\infty$. Exponentiating gives a limit of $0$, consistent with $\exp(t)\to0$ as $t\to-\infty$.

Worked Example: A power growing without bound

Find $\displaystyle\lim_{x\to\infty}x^{\sqrt{\ln x}}$.

For $x>1$, both the base and the exponent are positive, and the logarithmic expression is

q(x)=\sqrt{\ln x}\,\ln x=(\ln x)^{3/2}.

Because $\ln x\to+\infty$, it follows that $q(x)\to+\infty$. Therefore the original power satisfies

\lim_{x\to\infty}x^{\sqrt{\ln x}}=+\infty.

Check: The logarithm of the power grows without bound, and exponentiating a quantity that grows without bound also grows without bound.

What information does the logarithmic limit provide?

For a finite logarithmic limit $L$, the recovered limit $e^L$ is always positive. In particular, a finite logarithmic limit cannot produce a power limit of zero. Conversely, if a positive variable power tends to a positive finite number $A$, continuity of $\ln$ at $A>0$ implies that its logarithm tends to $\ln A$. This gives a useful two-way relationship for positive finite limits:

f(x)^{g(x)}\longrightarrow A>0 \quad\Longleftrightarrow\quad g(x)\ln f(x)\longrightarrow\ln A.

The forward implication uses continuity of $\ln$ at $A$; the reverse implication is the recovery theorem. For limits of zero or $+\infty$, the corresponding logarithmic behavior is $-\infty$ or $+\infty$, respectively, rather than a finite real number.

The theorem recovers a power limit only after the logarithmic limit has been established. If that calculation uses L'Hôpital's rule, its conditions still matter: the quotient must have an applicable indeterminate form, the functions must be differentiable on the relevant interval, and the denominator's derivative must be nonzero nearby. The recovery step does not remove those requirements.

A common mistake: omitting the exponent

A frequent error is to treat the logarithm of $f(x)^{g(x)}$ as just $\ln f(x)$. The correct identity is $\ln(f(x)^{g(x)})=g(x)\ln f(x)$ whenever $f(x)>0$. Omitting $g(x)$ can change the predicted limit completely.

For instance, consider the second worked example. The incorrect calculation would be

\ln\bigl((5+1/x)^{1/x}\bigr) \ \text{incorrectly replaced by}\ \ln(5+1/x)\longrightarrow\ln5,

leading to the wrong answer $e^{\ln5}=5$. The correction retains the exponent:

\ln\bigl((5+1/x)^{1/x}\bigr) = \frac{1}{x}\ln(5+1/x) \longrightarrow 0, \qquad \exp(0)=1.

Another important condition is positivity. If the base is not positive near the limiting point, the real logarithmic definition may not apply, so this method cannot automatically be used for a real-valued power.

A reliable recovery procedure

Check the base.
Confirm that $f(x)>0$ on the relevant domain near the limit point, so that $\ln f(x)$ is defined.

Form the logarithmic expression.
Set $q(x)=g(x)\ln f(x)$. Keep the exponent as a factor.

Find the limit of $q(x)$.
Use an appropriate limit method and check the conditions of any theorem you apply.

Exponentiate the result.
If $q(x)\to L$ for finite $L$, the power tends to $e^L$. If $q(x)\to-\infty$, it tends to $0$; if $q(x)\to+\infty$, it tends to $+\infty$.

The central idea is the identity $f(x)^{g(x)}=\exp(g(x)\ln f(x))$. Once the logarithmic expression has a limit, continuity or end behavior of the exponential recovers the power's limit. Keeping the base positive and the exponent inside the logarithmic product makes the method both precise and dependable.

Check Your Understanding

For each question, identify the logarithmic expression before deciding what the power does.

If $f(x)>0$ and $g(x)\ln f(x)\to\ln 4$, what is the limit of $f(x)^{g(x)}$?
If $g(x)\ln f(x)\to-\infty$, what does the power tend to?
Why must the base be positive to use the real logarithmic definition for a general real exponent?
For $x\to\infty$, determine the limit of $(3+1/x)^{2/x}$ by first finding its logarithmic expression.
State the error in replacing $\ln(f(x)^{g(x)})$ with $\ln f(x)$.

Recovering Power Limits from Logarithmic Limits

What You'll Learn

From a power to a logarithmic limit

The recovery theorem

Worked examples