# Applying l'Hospital's rule to find a limit

• October 16th 2011, 09:19 PM
skybluesea2010
Applying l'Hospital's rule to find a limit
I couldn't find the following limit:

limxx((1+1x)xe).
Definitely, it has to be through the L'Hospital's rule. We know that limx(1+(1/x))x=e. So, I wrote the above expression as
limx(1+1x)xe1/x.
Both numerator and denominator tend to zero as x tends to infinity. Here comes the L'Hospital's rule. I applied the L'Hospital's rule twice , but I got the limit equal to infinity after the second time which is clearly wrong. In the book, it says that the limit should be e/2.

Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your L'Hospital's rule to in both cases? Thanks
• October 16th 2011, 09:44 PM
Prove It
Re: Applying l'Hospital's rule to find a limit
Quote:

Originally Posted by skybluesea2010
I couldn't find the following limit:

limxx((1+1x)xe).
Definitely, it has to be through the L'Hospital's rule. We know that limx(1+(1/x))x=e. So, I wrote the above expression as
limx(1+1x)xe1/x.
Both numerator and denominator tend to zero as x tends to infinity. Here comes the L'Hospital's rule. I applied the L'Hospital's rule twice , but I got the limit equal to infinity after the second time which is clearly wrong. In the book, it says that the limit should be e/2.

Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your L'Hospital's rule to in both cases? Thanks

• October 16th 2011, 11:47 PM
CaptainBlack
Re: Applying l'Hospital's rule to find a limit
Quote:

Originally Posted by skybluesea2010
I couldn't find the following limit:

limxx((1+1x)xe).
Definitely, it has to be through the L'Hospital's rule. We know that limx(1+(1/x))x=e. So, I wrote the above expression as
limx(1+1x)xe1/x.
Both numerator and denominator tend to zero as x tends to infinity. Here comes the L'Hospital's rule. I applied the L'Hospital's rule twice , but I got the limit equal to infinity after the second time which is clearly wrong. In the book, it says that the limit should be e/2.

Any help please? Also, in case someone managed to solve it, please tell me which numerator and denominator did you apply your L'Hospital's rule to in both cases? Thanks

Presumably this means:

$\lim_{x\to \infty}\left\{ x\left[ \left(1+\frac{1}{x}\right)^x-e \right] \right\}$

CB
• October 17th 2011, 04:32 AM
skybluesea2010
Re: Applying l'Hospital's rule to find a limit
Exactly, this that's the expression i want to find its limit when x tends to infinity
• October 17th 2011, 06:23 AM
CaptainBlack
Re: Applying l'Hospital's rule to find a limit
Quote:

Originally Posted by CaptainBlack
Presumably this means:

$\lim_{x\to \infty}\left\{ x\left[ \left(1+\frac{1}{x}\right)^x-e \right] \right\}$

CB

Quote:

Originally Posted by skybluesea2010
Exactly, this that's the expression i want to find its limit when x tends to infinity

Then you apply L'Hopital's rule to:

$\lim_{x\to \infty} \frac{ \left(1+\frac{1}{x}\right)^x-e }{(\frac{1}{x})}$