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Someone told me if $\lim_{x\to{+\infty}} f(x)^{g(x)}=1^{+\infty}$ which is indeterminate limit then we can solve it by taking the following limit: $$k =\lim_{x\to +\infty}\big(f(x)-1\big)g(x)$$ So $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^k$$

I used this formula a lot and have seen it is fast practical way for this kind of limits, especially when we just want to know the value of the limit. What is the proof of this formula if it is true? Thanks.

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up vote 10 down vote accepted

If $$\lim_{x\to\infty}f(x)^{g(x)}=e^k\;,$$ then $$k=\ln\lim_{x\to\infty}f(x)^{g(x)}=\lim_{x\to\infty}\ln f(x)^{g(x)}=\lim_{x\to\infty}g(x)\ln f(x)\;.$$ Thus, in order for the method to work, we must have $$\lim_{x\to\infty}g(x)\ln f(x)=\lim_{x\to\infty}g(x)\left(f(x)-1\right)\;.$$

Using the Maclaurin series for $\ln(1+x)$, we have

$$\begin{align*} \ln f(x)&=\ln\left(1+f(x)-1\right)\\ &=\sum_{n\ge 1}(-1)^{n+1}\frac{\left(f(x)-1\right)^n}n\\ &\approx f(x)-1 \end{align*}$$

once $f(x)$ is close to $1$, and the error is bounded by $\frac12\left(f(x)-1\right)^2$, which decreases rapidly compared with $|f(x)-1|$.

You can also compare what happens when you apply l’Hospital’s rule to $$\lim_{x\to\infty}\frac{\ln f(x)}{1/g(x)}$$ and to $$\lim_{x\to\infty}\frac{f(x)-1}{1/g(x)}\;:$$ the former gives you $$\lim_{x\to\infty}\frac{f\,'(x)/f(x)}{-\left(g(x)\right)^{-2}g'(x)}=\lim_{x\to\infty}\frac1{f(x)}\lim_{x\to\infty}\frac{f\,'(x)}{-\left(g(x)\right)^{-2}g'(x)}=\lim_{x\to\infty}\frac{f\,'(x)}{-\left(g(x)\right)^{-2}g'(x)}\;,$$ which is exactly what the latter gives you.

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Dear Brian, what happens when we have $c^{+\infty}$ instead of $1^{+\infty}$. Will the limit be changed? $c$ is constant.Thanks – Babak S. Sep 13 '12 at 7:20
Then the limit is no longer an indeterminate form. If $0\le c<1$, the limit is $0$, and if $c>1$, it’s $+\infty$. – Brian M. Scott Sep 13 '12 at 7:33
You won't believe if you know how many indeterminate limits I have evaluated till now by your result. Thanks dear Brian. :-) – Babak S. Jan 5 '13 at 17:51
@Babak: You’re very welcome! – Brian M. Scott Jan 5 '13 at 17:56

This is easy to understand if we note that the fundamental limit concerning the logarithmic function $$\lim_{y \to 0}\frac{\log(1 + y)}{y} = 1$$ In the particular case here we have $\lim_{x \to \infty} f(x) = 1$ and $\lim_{x \to \infty} g(x) = \infty$ and if the following limit $$L = \lim_{x \to \infty}\{f(x)\}^{g(x)}$$ exists then we have

$\displaystyle \begin{aligned}\log L &= \log\lim_{x \to \infty}\{f(x)\}^{g(x)}\\ &= \lim_{x \to \infty}\log\{f(x)\}^{g(x)}\text{ (because of continuity of log)}\\ &= \lim_{x \to \infty}g(x)\log f(x)\\ &= \lim_{x \to \infty} g(x)\log \{1 + f(x) - 1\}\\ &= \lim_{x \to \infty} g(x)\{f(x) - 1\}\frac{\log \{1 + f(x) - 1\}}{f(x) - 1}\\ &= \lim_{x \to \infty} \{f(x) - 1\}g(x)\lim_{y \to 0}\frac{\log(1 + y)}{y}\text{ (put }y = f(x) - 1)\\ &= \lim_{x \to \infty}\{f(x) - 1\}g(x)\end{aligned}$

There is no need to use series expansions or L'Hospital's rule in most of the limit problems if we make proper use of existing limit formulas. I still prefer to use the logarithmic limit directly rather than relying on special formulas (as mentioned in this question) for each special indeterminate form. Such formulas are good only when someone is appearing in objective (multiple choice type) competitive exams and these formulas do not help an iota for a beginner learning limits. Unfortunately in my country most competitive exams have become multiple choice type.

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