If $\lim_{x\to a}f(x)=1$ and $\lim_{x\to a}g(x)=\infty$ then show that $$\lim_{x\to a}\{f(x)\}^{g(x)}=e^{\lim\limits_{x\to a}{g(x)\{f(x)-1\}}}.$$ I started off as:
$$\lim_{x\to a}\{f(x)\}^{g(x)} = e^{\lim\limits_{x\to a}{g(x)\{\ln f(x)\}}}$$ $$=e^{\lim\limits_{x\to a}{\{\ln f(x)\}\over {\frac{1}{g(x)}}}}$$ $$=e^{-\lim\limits_{x\to a}{\frac{f'(x)}{g'(x)}.\frac{\{g(x)\}^2}{f(x)}}}.$$
But this doesn't seem to be going anywhere.
Hint: use $\lim\limits_{x \to 0}\frac{\ln (1+x)}{x}=1$.
Then $\lim\limits_{x \to a} \ln f(x) = \lim\limits_{x \to a} \left( \frac{\ln (1+[f(x)-1])}{f(x)-1}\cdot[f(x)-1] \right)=...?$
$$\begin{align} \lim_{x\rightarrow\infty}f(x)^{g(x)}&=\lim_{x\rightarrow\infty}\left[1+\frac1{\frac1{f(x)-1}}\right]^{g(x)}\\ &=\lim_{x\rightarrow\infty}\left[1+\frac1{\frac1{f(x)-1}}\right]^{\frac1{f(x)-1} \cdot(f(x)-1)g(x)}\\ &=\lim_{x\rightarrow\infty} e^{(f(x)-1)g(x)}\phantom{\frac1{\frac1n}} \end{align}$$
If
$\lim_{x→α}f(x)=1$ and $\lim_{x→α}g(x)= \infty $
then $$\lim_{x→α}(f(x))^{g(x)}= \lim_{x→α}(1+f(x)−1)^{g(x)}= \lim_{x→α}[[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{(f(x)-1)}]^{g(x)}= \lim_{x→α}[(1+f(x)−1)^{\frac{1}{f(x)-1}}]^{\lim_{x→α}(f(x)-1)g(x)}= e^{\lim_{x→α}(f(x)-1)g(x)}.$$
\phantom{blah}produces a blank area of the same size as its argument would take up if typeset. Normally, that's used for horizontal alignment but I used it in your example to give a bit more vertical space. – David Richerby Apr 26 '14 at 12:41\vphantomwould have been a better way but, hey,\phantomworked. – David Richerby Apr 29 '14 at 08:23