Id highly appreciate it if anyone could help me with any of the following
problems!

What is the antiderivative of (cosx)^4 (cos(x^4))??

Find lim as x->infinity [(2x-3)/(2x+5)]^2x+1

Find lim as x->infinity [e^(x^2) - e^ [(x^2) + xsinx)] ]

Find f ' (x) and f '' (x) if f(x)= e^(-1/sin^2x) ,where x is between negative pi/2 and positive pi/2
f(x)= 0 , where x = 0

2. Originally Posted by shmayer
Id highly appreciate it if anyone could help me with any of the following
problems!

What is the antiderivative of (cosx)^4 (cos(x^4))??

Find lim as x->infinity [(2x-3)/(2x+5)]^2x+1

Find lim as x->infinity [e^(x^2) - e^ [(x^2) + xsinx)] ]

Find f ' (x) and f '' (x) if f(x)= e^(-1/sin^2x) ,where x is between negative pi/2 and positive pi/2
f(x)= 0 , where x = 0

$\int\cos^4(x)\cos(x^4)\,dx$ has no closed from solution.

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There might be a slick way to do this, but here's a method that will work, though it's a bit tedious and may cause retinal bleeding.

$\lim_{x\to\infty}\left(\frac{2x-3}{2x+5}\right)^{2x+1}=\lim_{x\to\infty}\exp\left[(2x+1)\ln\left(\frac{2x-3}{2x+5}\right)\right]=$ $\exp\left[\lim_{x\to\infty}\frac{\ln\left(\frac{2x-3}{2x+5}\right)}{(2x+1)^{-1}}\right]$

Use L'Hopital's rule on that limit.

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$\lim_{x\to\infty}e^{x^2}-e^{x^2+x\sin x}$

This limit is undefined, because $e^{x^2}-e^{x^2+x\sin x}=e^{x^2}\left(1-e^{x\sin x}\right)$. Since $\lim_{x\to\infty}x\sin x$ DNE, the whole limit DNE.

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$f(x)=\exp\left(\frac{-1}{\sin^2x}\right)$

For these derivatives, just use the chain rule.

3. hey, thanks for the reply..for the one question you said it DNE since limx->infinity (xsinx) DNE...but couldnt i use Lhopitales rule before you factored out e^x ?

4. L'Hopital's rule requires $\frac{0}{0}$ or $\frac{\infty}{\infty}$. We didn't have either of those.

5. Cant we use lhopitales rule for infinity - infinity? that is indeterminate form right?

6. Originally Posted by shmayer
Cant we use lhopitales rule for infinity - infinity? that is indeterminate form right?
No, it doesn't work for that form. For example, consider:

$\lim_{x\to\infty}(x-\ln x)$

This is $\infty-\infty$, and the whole limit equals $\infty$.

But if you try to use L'Hopital's rule, you get $\lim_{x\to\infty}\left(1-\frac{1}{x}\right)=1$, which is wrong.

7. Ah i see thanks! Also, why does the first question not have a solution?

8. There's no solution that can be expressed in terms of elementary functions whose derivative is $\cos(x^4)\cos^4(x)$.

9. Originally Posted by redsoxfan325
There's no solution that can be expressed in terms of elementary functions whose derivative is $\cos(x^4)\cos^4(x)$.
Why is this the case?

10. also, how would i differentiate f and find the domain of f(x)= lnlnlnx ?

Thanks a TON!

11. Originally Posted by shmayer
Why is this the case?
Because none of the methods used to integrate (u-sub, by parts, etc.) yield anything integrable. I know that's a pretty lousy answer; maybe someone else on MHF can explain it better.

Originally Posted by shmayer
also, how would i differentiate f and find the domain of f(x)= lnlnlnx ?

Thanks a TON!
For differentiation, you have to use the chain rule.

For the domain, $\ln(u)$ is defined for $u>0$. In this case, $u=\ln(\ln(x))$, so we need $\ln(\ln(x))>0$. This is the case when $\ln(x)>1$. (Do you see why?) And $\ln(x)>1$ when $x>e$. So the domain of $\ln(\ln(\ln(x)))$ is $(e,\infty)$.

12. Thanks!