# Difficult Limit

• Oct 15th 2012, 05:06 PM
DylanRenke
Difficult Limit
Limit as x goes towards 0 for the function (e^x-1)^2/(sin(e^x-1))

Any help would be much appreciated
• Oct 15th 2012, 06:20 PM
TheEmptySet
Re: Difficult Limit
Quote:

Originally Posted by DylanRenke
Limit as x goes towards 0 for the function (e^x-1)^2/(sin(e^x-1))

Any help would be much appreciated

To simplify things let

$\displaystyle u=e^x-1$

If we take the limit as $\displaystyle x \to 0$ this is the same as the limit as $\displaystyle u \to 0$

Now this gives you the limit

$\displaystyle \lim_{x \to 0}\frac{(e^{x}-1)^2}{\sin(e^x-1)}=\lim_{u \to 0}\frac{u^2}{\sin(u)}$

This gives

$\displaystyle \lim_{u \to 0}\frac{u}{\sin(u)}\cdot u=1 \cdot 0 =0$
• Oct 15th 2012, 07:33 PM
DylanRenke
Re: Difficult Limit
Why doesnt that last line just give you 0/0? because when u approaches 0, it equals 0, and when sin(u) approaches 0, its zero as well?
• Oct 15th 2012, 07:41 PM
TheEmptySet
Re: Difficult Limit
Quote:

Originally Posted by DylanRenke
Why doesnt that last line just give you 0/0? because when u approaches 0, it equals 0, and when sin(u) approaches 0, its zero as well?

This is a well known limit and its value should be known. The proof is a bit lengthy and requires a bit of trigonometry and the use of the squeeze theorem.

Do you need a proof?
• Oct 15th 2012, 07:53 PM
DylanRenke
Re: Difficult Limit
It would be quiet helpful, because i really do not understand it.
• Oct 15th 2012, 07:55 PM
DylanRenke
Re: Difficult Limit
Actually, i have found that the professor told us to memorize that limit. Thank you for the help