Why is $\lim\limits_{x\to\infty} e^{\ln(y)} = e^{\,\lim\limits_{x\to\infty} \ln(y)}$?

Question

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

Answer 1

It is true for continuous functions that when $\lim_{n\to\infty} z_n = z$ exists (so $z$ finite), that $$\lim_{n\to\infty} f(z_n) = f(\lim_{n\to\infty} z_n)$$

In the problem above we can let $z_n = \log(x^n)^{1/x_n}$ for some sequence $x_n \to \infty$. In that case, $$\lim_{n\to\infty} z_n = 0$$

Now since $e^x$ is a continuous function we have: $$\lim_{n \to \infty} \exp{(\log(x_n)^{1/x_n})} = \exp(\lim_{n\to\infty}\log(x_n)^{1/x_n}) = 1$$

Answer 2

As exponential is a continuous function we can pass the limits inside the function

Lemma: If a sequence $(x_n)_{n \in \mathbb{N}} \to x$ and we have a continuous function f then $\lim_{n \to \infty}f(x_n)=f(\lim_{n \to \infty} x_n)=f(x)$ And in your case $f(x)=e^x$ But the for this to be true the limits needs to exist

Answer 3

This question has been already answered here:

Q: I was studying L'Hopital's rule and how to deal with indeterminate forms of the type 0^0. It's not clear to me how lim e^f(x) = e^lim f(x).

A: You can move the limit inside the exponential, because the exponential itself doesn't have problem spots ("is continuous everywhere"), so it is only the f(x) inside that you have to deal with regarding the limit.

Reference https://www.physicsforums.com/threads/limits-involving-exponential-functions.330771/

Answer 4

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.