L'hospitals rulee major help

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 plzz

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

The Attempt at a Solution

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Hint

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

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

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

noo..havent learned it =S

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..

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?

noo i haven't :|

kk thnnxx for your help newayzz =)

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

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

Dick, Brilliant!

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