Find limit using L'Hopital's rule, ln, and e: how did they do these steps?

[Nishiura_high]

Homework Statement

This is from an online answer, and I don't understand the steps that it took. How did they go from the first red box to the second red box?

Homework Equations

L'Hopital's rule
Laws of exponents

The Attempt at a Solution

I am really really confused. It looks like they took what e was raised to, and found the limit of that... but what rule allowed them to "Finally, plug in the original limit"?

 

[lurflurf]

So the step is

\lim_{x\rightarrow\infty} e^{x\log(1+1/x^2)}=\exp\left( \lim_{x\rightarrow\infty} \left( x\log\left( 1+\frac{1}{x^2} \right) \right) \right)

Which is justified by the continuity of the exponential. The general rule is for f continuous

\lim_{x\rightarrow\infty} f(g(x))=f \left( \lim_{x\rightarrow\infty} g(x) \right)

This is sometimes called the composition rule for limits.

Then to keep out expression from being messy we can substitute
call the limit a, so we want to find e^a
we can find a first, then find e^a

 

[pierce15]

First, you need to determine whether or not this is an indeterminate form. As x tends to infinity, the base comes closer to one, and the exponent naturally increases to infinity. Since 1^\infty is an indeterminate form, you can use L'Hopital's rule here. The first step is to use the property e^{ln(x)}=x This is a very useful property because the logarithm will turn this limit into an indeterminate product, which is much easier to work with. So back to the problem:
\lim_{x\to\infty} e^{ln((1+\frac{1}{x^2})^x)}
\lim_{x\to\infty} e^{xln(1+\frac{1}{x^2})}
Once again, we have a product of two infinities, so this is another indeterminate form. We can rewrite this:
\lim_{x\to\infty} e^\frac{ln(1+\frac{1}{x^2})}{\frac{1}{x}}
This looks pretty nasty, but it's doable. Can you use L'Hopital's rule now?