Thread: Difficult Limit

1. 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

2. Re: Difficult Limit

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

$\displaystyle \lim_{u \to 0}\frac{u}{\sin(u)}\cdot u=1 \cdot 0 =0$

3. 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?

4. Re: Difficult Limit

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?

5. Re: Difficult Limit

It would be quiet helpful, because i really do not understand it.

6. Re: Difficult Limit

Actually, i have found that the professor told us to memorize that limit. Thank you for the help