# Limits

• Jul 19th 2011, 02:00 AM
Arron
Limits
I have worked out the following but need to prove it.

lim
x-->0

x^2
1-e^x^2

= -1

Can anyone help?
• Jul 19th 2011, 02:03 AM
Prove It
Re: Limits
It's of the indeterminate form $\displaystyle \frac{0}{0}$ so you can use L'Hospital's Rule.
• Jul 19th 2011, 02:15 AM
CaptainBlack
Re: Limits
Quote:

Originally Posted by Arron
I have worked out the following but need to prove it.

lim
x-->0

x^2
1-e^x^2

= -1

Can anyone help?

For small $x$

$e^{x^2}=1-x^2+R(x)$

where $|R(x)|\le |x|^4/2$

So:

$\frac{x^2}{1-e^{x^2}}=\frac{1}{-1+[R(x)/x^2]}$

etc.

CB
• Jul 19th 2011, 04:22 AM
Jester
Re: Limits
If you let $t = x^2$, you could write your limit as

$\displaystyle - \dfrac{1}{\lim_{t \to 0} \dfrac{e^t-1}{t}}$

with the limit now something you might recognize.