L'hospitals rulee major help

Hint

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

 

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

 

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?

 

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.

 

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.

 

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

[tex] \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} [/tex]

 

Dick, Brilliant!

 

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.

 

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

 

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