# Finding a limit

1. Nov 15, 2008

### squeetox

1. The problem statement, all variables and given/known data
I need to find the following limit.

2. Relevant equations
$$\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)}$$

3. The attempt at a solution
I think it's got to be something with Taylor series, but I don't really know how to do it.

2. Nov 15, 2008

### tiny-tim

Welcome to PF!

Hi squeetox! Welcome to PF!

Hint: for example, the Taylor series for e3x is 1 + 3x + … , so if (e3x - 1) was a factor, you could replace it (as x -> 0) by 3x.

3. Nov 15, 2008

### squeetox

Thanks!

Why do the others terms (9x2/2 + 9x3/2...) disappear as x->0?

PS: I had thought of replacing $\sin^3 x (e^{x^2}) \sim x^5$ because I know that $\sin [f(x)] \sim f(x)$ and $e^{f(x)} -1 \sim f(x)$. The rest I have no clue though.

Last edited: Nov 15, 2008
4. Nov 16, 2008

### tiny-tim

Hi squeetox!
They don't actually disappear … but, compared with x, they get so small (as x -> 0) that we can ignore them.
Nooo … you're confusing ex2 and ex2 - 1.

5. Nov 16, 2008

### springo

So basically you mean that since it's a sum of x's to different powers, and the ones after 3x get smaller much faster, I can ignore them?
Thanks.

6. Nov 16, 2008

### squeetox

I'm sorry that's what I meant. If you look at the limit I'm looking for, it's $e^{x^2} - 1 \sim x^2$ therefore $sin^3 x(e^{x^2}-1) \sim x^5$.

Now what about $(x-\sinh x)(\cosh x- \cos x)$ and $5+\sin x \ln x$?

7. Nov 16, 2008

### tiny-tim

ah yes, so it is … i didn't look back up to the top!
The first lot, you do the same way.

The sinx lnx … I can't remember whether there's a Taylor expansion for lnx … it obviously does -> 0 … just look at sin10-100 ln10-100, but how to prove it?

anyway, that's you problem, not mine!

8. Nov 16, 2008

### squeetox

For sin x ln x, I'm thinking...
$$\sin x \ln x \sim x \ln x$$

And
$$\lim_{x\rightarrow0}x \ln x = \lim_{x\rightarrow\infty} \frac{1}{x}\ln\frac{1}{x} = \lim_{x\rightarrow\infty} -\frac{\ln x}{x} = 0$$

Therefore:
$$\lim_{x\rightarrow0}(5+\sin x \ln x) = 5$$

For the first lot, do I find Taylor expressions for (x - sinh x) and (cosh x - cos x) or for the product of both?

Edit: I got it:
$$x-\sinh x \sim -\frac{x^3}{6}$$
$$\cosh x - \cos x \sim x^2$$

So finally...
$$\lim_{x\rightarrow0}\frac{(x-\sinh x)(\cosh x- \cos x)}{(5+\sin x \ln x) \sin^3 x (e^{x^2}-1)} = -\frac{1}{30}$$

Thank you so much for your help!

Last edited: Nov 16, 2008