- Fri Jan 09, 2009 9:25 pm
- Forum: Calculus
- Topic: Limit approaching infinity with an unknown in the exponent
- Replies:
**7** - Views:
**17388**

That's what I'm talking about! Thanks. I follow the L'Hopital's Rule method of solving this problem, but where did you get the " \lim_{n\to\infty}\left(1+\frac1n\right)^n = e " from? This is a given identity thinger I should be familiar with? I haven't come across it, please let me...

- Mon Dec 22, 2008 7:57 pm
- Forum: Calculus
- Topic: Limit approaching infinity with an unknown in the exponent
- Replies:
**7** - Views:
**17388**

That's not helping me dave. It looks good intuitively but 1 is not a valid answer. I repeat, one raised to infinity is considered an indeterminate form. Maybe someone else would like to chime in?

- Sat Dec 20, 2008 2:55 pm
- Forum: Calculus
- Topic: Limit approaching infinity with an unknown in the exponent
- Replies:
**7** - Views:
**17388**

2 divided by infinity goes to 0. 1 plus 0 is one. One raised to infinity is an indeterminate form, hence my wish to change the nature of the exponent *x*.

- Thu Dec 18, 2008 3:27 pm
- Forum: Calculus
- Topic: Limit approaching infinity with an unknown in the exponent
- Replies:
**7** - Views:
**17388**

Find the limit, as x approaches infinity, of (1+(2/x)) x This is a problem from Temple University's 'Calculus on the Web' site, I am trying to get re-acquainted with first-semester Calculus as I will begin taking Calc II in a month from now. I can't think of how to bring that x down. Logs? Thanks, J...