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

In the above limit $y = x ^{\frac 1x}$. Is the above a limit or an exponent property?

Context (Last paragraph): http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

• Your equation is not true for the standard real numbers. The left hand side is $\lim_{x\to\infty}x$ which is undefined, and the right hand side has an undefined limit in the exponent. If you use the affine extended real number system, you could say both sides equal $+\infty$: Is that your question? – Rory Daulton Jun 5 '15 at 20:58
• @RoryDaulton "lim x as x→∞ which is undefined" this is false, "lim of x as x→∞ equals ∞ – Jane Smith Jun 5 '15 at 21:06
• Doy you mean $x=\infty$, not a smth like $x=a$? – Michael Galuza Jun 5 '15 at 21:06
• You cannot blame others for editing your question. It was not clear what that $y$ was before you provide the link. – user99914 Jun 5 '15 at 21:16
• Apologies, I thought you must have meant $x$ inside the $\ln$. – muaddib Jun 5 '15 at 21:17

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: $$\lim_{n \to \infty} \exp{(\log(x_n)^{1/x_n})} = \exp(\lim_{n\to\infty}\log(x_n)^{1/x_n}) = 1$$

• I'd say this is only half true. Suppose he had e ^ - ln(x). then lim(x->inf)(e^(-ln(x)) = 0, but -ln(x) itself has no limit (it goes to minus infinity) – Dleep Jun 5 '15 at 20:40

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.

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.

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

• Is that still true for "$x=\infty$"? (BTW, the question is changed) – user99914 Jun 5 '15 at 21:10
• @John For the limit to exist it has to converge to a value in $\mathbb{R}$ .So indeed it's not valid if we are not in the extended real number system. – user3503589 Jun 5 '15 at 21:15