TITLE: Suppose given that $ \lim_{x \to 0}f(x)=1 \lim_{x \to 0}g(x)=\infty \lim_{x \to 0}g(x)(f(x)-1)=c $
QUESTION [6 upvotes]: Suppose given that 
$$ \lim_{x \to 0}f(x)=1 \\ \lim_{x \to 0}g(x)=\infty \\ \lim_{x \to 0}g(x)(f(x)-1)=c  $$
Then find the value of $ \lim_{x \to 0} f(x)^{g(x)} $
So this is what I've done. Please verify if its correct approach @Macavity
$$ f(x)^{g(x)} =  e^{\log{f(x)^{g(x)}}} = e^{{g(x)\log f(x)}}  $$  $(1)$
Now from Taylor series expansion of $\log(1+x)$ Substitute $x$ by $f(x)-1 $ in the Taylor series
$$\log f(x)=  (f(x)-1) - \frac12(f(x)-1)^2 \cdots$$    $(2)$
From $(1)$ and $(2)$ By taking only the first term of the $\log f(x) $ expansion 
$$ \lim_{x \to 0} f(x)^{g(x)}= e^c  $$
Is this approach correct?

REPLY [2 votes]: Your approach is intuitively correct. But a simpler approach with rigorous justification of all the steps is as follows.
Let $L = \lim_{x \to 0}f(x)^{g(x)}$ so that
\begin{align}\log L &= \log\left(\lim_{x \to 0}f(x)^{g(x)}\right)\notag\\
&=\lim_{x \to 0}\log f(x)^{g(x)}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0}g(x)\log f(x)\notag\\
&= \lim_{x \to 0}g(x)\log \{1 + f(x) - 1\}\notag\\
&= \lim_{x \to 0}g(x)(f(x) - 1)\dfrac{\log \{1 + f(x) - 1\}}{f(x) - 1}\notag\\
&= \lim_{x \to 0}g(x)(f(x) - 1)\cdot\lim_{t \to 0}\dfrac{\log (1 + t)}{t}\text{ (by putting }t = f(x) - 1)\notag\\
&= c\cdot 1 = c\end{align} and hence $L = e^{c}$. We have used the standard limit $$\lim_{t \to 0}\frac{\log(1 + t)}{t} = 1$$ Also note that the second condition $\lim_{x \to 0}g(x) = \infty$ is unnecessary as can be seen in the above derivation.