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

  • #1

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?

screenshot2.png



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"?
 

Answers and Replies

  • #2
lurflurf
Homework Helper
2,423
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So the step is

[tex]\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)[/tex]

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

[tex]\lim_{x\rightarrow\infty} f(g(x))=f \left( \lim_{x\rightarrow\infty} g(x) \right)[/tex]

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
 
Last edited:
  • #3
315
2
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 [tex]1^\infty[/tex] is an indeterminate form, you can use L'Hopital's rule here. The first step is to use the property [tex]e^{ln(x)}=x[/tex] 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:
[tex]\lim_{x\to\infty} e^{ln((1+\frac{1}{x^2})^x)}[/tex]
[tex]\lim_{x\to\infty} e^{xln(1+\frac{1}{x^2})}[/tex]
Once again, we have a product of two infinities, so this is another indeterminate form. We can rewrite this:
[tex]\lim_{x\to\infty} e^\frac{ln(1+\frac{1}{x^2})}{\frac{1}{x}}[/tex]
This looks pretty nasty, but it's doable. Can you use L'Hopital's rule now?
 
Last edited:

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