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meg24209 · Aug 27, 2007, 07:36 PM

what is the limit as x-->0 when tan^2x / x

shygrneyzs · Aug 27, 2007, 07:42 PM

We do not do homework for people on the site. We do offer helps, can give advice on your work, but we are not going to give the answer.

I will give you a website on math.
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meg24209 · Aug 27, 2007, 07:44 PM

I am not trying to get someone to do my hw!! I have worked hard on this problem and I cannot figure it out. I was just looking for some help. Even if someone could me the first step! Not the answer!! Shyrgrneyzs

galactus · Aug 28, 2007, 03:44 AM

Try rewrting as :

$\lim_{x\to\0}\left(\frac{sin^{2}(x)}{x}\cdot\frac{ 1}{cos^{2}(x)}\right)$

meg24209 · Aug 28, 2007, 07:15 AM

That mean sinx/x--> 1 because of properties of limits. And that leaves sinx(1/cos^2x)

galactus · Aug 28, 2007, 08:39 AM

Another thing you could do is use L'Hopital.

Find the derivative of $tan^{2}x$

That is $2sec^{2}(x)tan(x)$

The derivative of x is 1, so you have:

$2\lim_{x\to\0}sec^{2}(x)tan(x)$

Is that easier?

BTW, $\lim_{x\to\0}\frac{sin^{2}x}{x}=0$ , not 1.

meg24209 · Aug 28, 2007, 03:24 PM

I didn't say that. I said sinx/x is 1.

My teacher helped me solve the problem and gave the answer 0.
Thanks for the help.

galactus · Aug 28, 2007, 06:32 PM

I know you didn't say that. I was pointing it out from my previous post. Yes, the answer is 0.