evaluate lim as x goes to infinity (x/(x+1))^x?

Question

tip, l'hopital's rule is required

Answer 1

When you are trying to evaluate a limit that has a power like this you normally want to take the natural log first. Doing that in this problem you get:

x*log(x/(x+1))

Now you generally want to apply l'Hopital's rule. To do this you want to put the limit in the form 0/0 or infinity/infinity. Once you use l'Hopital's rule to evaluate this, you exponentiate to get back to the original limit (since you took the log to start out).

Answer 2

You don't need L'Hopital's rule. As far as I can tell, that just makes things hopelessly messier.

Limit_{x->infty}(x^x/(x+1)^x)
=(1+1/x)^-x (I divided both numerator and denominator by x^x)

The definition of e (natural exponent)
e=limit_{n->infty}(1+1/n)^n

So the limit here is 1/e

Now, I thought n was discrete, and x being continuous, that might be issue, but it's not. Looking up "e" at mathworld.com gives this exactly limit, with the x's, and Mathematica checks this result as well.

Edit: I stand corrected. The above poster's method works, but there's some nasty algebra involved. Taking the log of the limit is also slightly dangerous because it has some requirements behind it.

If you want to do it that way, take the log and be very careful about the derivative and the resulting algebraic simplification.

After all of it, you should get the limit = -1. Of course, exponentiating this gives you e^-1, which is what I got before. Sorry!