MHB Limit involving exponential function

[Bueno]

Hello everyone, how are you?

I'm having trouble to evalue the following limit:

$$\lim_{x->\infty} (\frac{x}{1+x^2})^x $$

I "transformed" it into $$e^{ln{(\frac{x}{1+x^2})^x}}$$ and tried to solve this limit:

$$\lim_{x->\infty} x ln{(\frac{x}{1+x^2})}$$

But I have no idea how to solve it correctly. Can you help me?

Thank you,

Bueno

 

[Jameson]

I haven't solved this either, but want to post my thoughts until someone else can help. I did the same thing you did to begin.

Let $$L= \lim_{x \rightarrow \infty}\left( \frac{x}{1+x^2} \right)^x$$

$$\ln(L) = \lim_{x \rightarrow \infty}x \ln\left( \frac{x}{1+x^2} \right) = \lim_{x \rightarrow \infty} x \left[ \ln(x)-\ln(1+x^2) \right]= \lim_{x \rightarrow \infty} x\ln(x)-x\ln(1+x^2)$$

This seems good so far and appropriate now to use L'Hopital's Rule, but I see one problem. The answer to this limit is $0$, so if $L=0$ then $\ln(L)$ isn't defined and I don't see how we can get the answer this way, so it seems like another method should be used. This limit uses a trick I haven't seen before or forgot, because it's tougher than many I have been browsing through just now.

Will post back if I see the solution and I hope that someone helps you soon! :)

 

[MarkFL]

We may rewrite the expression as:

$$\left(\frac{\frac{1}{x}}{1+\frac{1}{x^2}} \right)^x$$

and we see we have the form:

$$0^{\infty}$$

which is not indeterminate, and is in fact equal to zero.

 

[Jameson]

Nicely done, Mark. (Clapping). I knew that the indeterminate form method wasn't going to work and thought of rewriting the expression somehow but just didn't see it.

 

[Bueno]

I couldn't see anything like this, it worked perfectly. It's the first limit I see that needs this kind of trick.
Thank you!

I've been studying this kind of limits today and most of them were solved by the technique I mentioned in the first post, except for the one you just showed me how to solve and another.

Bueno