TITLE: Why is $\lim\limits_{x\to\infty} e^{\ln(y)} = e^{\,\lim\limits_{x\to\infty} \ln(y)}$?
QUESTION [5 upvotes]: In the above limit $y = x ^{\frac 1x}$. Is the above a limit or an exponent property?
Thanks in advance.
Context (Last paragraph): http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

REPLY [2 votes]: The theorem cited by muaddib is valid when n is natural. Nevertheless, in your question, you want that x be real. And Jane Smith's aswer makes senses, but it's pretty qualitativy. I will try to prove the following statement:
If $f$ is continuous in an interval I$\in \mathbb{R}$ and $L \in \mathbb{}$ I, then $$\ lim_{x\to\ p}f(g(x))= f(lim_{x\to\ p}g(x))$$ $$\ lim_{x\to\infty}f(g(x))= f(lim_{x\to\infty }g(x))$$
Since the limits of g at right member of equation exist.
We can do $lim_{x\to\ p}g(x)=L$, then for all $\epsilon >0$, we have a $\delta$ such that $$|x-p|<\delta => |g(x)-L|<\epsilon$$
Since $f$ is continuous, then for all $\epsilon_1 >0$, we have a $\delta_1$
$$|g(x)-L|<\delta_1 => |f(g(x))-f(L)|<\epsilon_1$$
It's enough makes $\delta_1=\epsilon$. We can do that because $\epsilon$ is arbitrary, so it can assume any real positve value. In the second case $x \to \infty$, it's suffcient you switch $p$ for $\infty$ and do $x>\delta$. The rest is the same. The case $x \to -\infty$ is analogous to case $x \to \infty$.
If $|lim_{x\to p}g(x)|= \infty$, then for all $\epsilon >0$, we have a $\delta$ such that $$|x-p|<\delta => |g(x)|>\epsilon$$
Now we have to investigate the behavior of $f(y)$ when $|lim_{x\to p} y|= \infty$. In case $|lim_{x\to \infty}g(x)|= \infty$, it's analogous. Therefore, in any cases above, you only have to investigate the limit of g.