How can I solve limx->infinity (x/x+1)^x using L'Hospital's rule?

[angel_eyez]

l'hospitals rulee! major help

1. i don't know how to do this question
limx->infinity (x/x+1)^x. i keep gettin the rong answer, some heelp pleasez



2.the right answer is 1/e but I am not getting that, i tried many diff ways, i can't post my work it here because its too long and confusing

The Attempt at a Solution

 

[christianjb]

Hint

Consider the series expansion for e^[1/x], when x is large.

 

[angel_eyez]

umm so when x is large 1/x = 0 so that means e^o = 1, I am still lossttt

 

[christianjb]

Do you know the Taylor series for e^y, when y is small?

 

[angel_eyez]

noo..havent learned it =S

 

[angel_eyez]

i took the ln of both sides, and then found the derivative, and i got the answer as zero.. so i got ln y= 0
den i used the fact that e^lny = y...there fore y= e^0 =S..

 

[christianjb]

Hmmm, I'm not sure how to do it without recourse to a Taylor series expansion. It's easy to prove though that

e^x=1+x+x^2/2!+x^3/3+...

Have you done Taylor series at all?

 

[angel_eyez]

noo i haven't :|

 

[angel_eyez]

kk thnnxx for your help newayzz =)

 

[christianjb]

You suggested that L'hopital's rule should be used in the derivation. I played around with that and couldn't see it, but maybe someone else can work it out.

Let me know when you get the answer.

 

[Dick]

You should have gotten something like x*ln(x/(x+1)) when you took the log. This is an infinity*0 form. To use L'Hopital you want to write it as 0/0 (or infinity/infinity). So write it as ln(x/(x+1))/(1/x). Now take derivative of the numerator and denominator and send x->infinity.

 

[dextercioby]

There's no need for l'Hospital's rule, as this limit is elementary

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

 

[christianjb]

Dick, Brilliant!

 

[christianjb]

There's no need for l'Hospital's rule, as this limit is elementary

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

That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.

 

[Dick]

Dick, Brilliant!

Thanks. Wish I could say I invented it, but it's a pretty routine thing to do.

 

[dextercioby]

That's supposing that you know the Taylor series expansion of e^x. It's nice that you can prove it a completely different way.

Pardon me ? I just the definition of "e". What's more elementary than that ??