# Lim x->0 (arctan(x^2))/(sinx+tanx)^2

1. Dec 1, 2011

### skyturnred

1. The problem statement, all variables and given/known data

lim x→0 $\frac{arctan(x^{2})}{(sinx+tanx)^{2}}$

2. Relevant equations

3. The attempt at a solution

OK, so at first, the form is $\frac{0}{0}$ so I use L'Hopital's rule. I get the following as my answer.

lim x→0 $\frac{1}{1+x^{4}}$*$\frac{1}{2(sinx+tanx)(cosx+sec^{2}x}$

But the form is not 0/0 and I cannot figure out what to do at this point. I have tried expanding out the (sinx+tanx) with the (cosx+sec^2x) to no avail.. and I can't really think of what else to do.

2. Dec 1, 2011

3. Dec 1, 2011

### Dick

You did not differentiate arctan(x^2) correctly. Don't forget the chain rule!

4. Dec 1, 2011

### skyturnred

Thanks! But now I have another issue.

When I fix my error, I get

lim x→0 $\frac{x}{(1+x^4)(sinx+tanx)(cosx+sec^2x)}$

This is of the form 0/0

I tried to expand the bottom and I get a HUGE equation, a very very long equation and I must be doing this wrong. Any ideas?

PS: This is the equation I get.

$\frac{x}{sinxcosx+sinxsec^{2}x+sinx+tanxsec^{2}x+x^{4}sinxcosx+x^{4}sinxsec^{2}x+x^{4}sinx+x^{4}tanxsec^{2}x}$

I really doubt that I am expected to use hopitals rule on that.. even though it wouldn't be too hard

5. Dec 1, 2011

### Dick

What's lim x->0 of 1/((1+x^4)*(cos(x)+sec(x)^2))? That part is not indeterminant. It just approaches a constant. Just do l'Hopital on the indeterminant part.

6. Dec 1, 2011

### skyturnred

Oh, I completely forgot you could do that! I get 1/4 is the answer to the question.. which I hope is right. And thanks so much for your help!