View Full Version : Limits!
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.
QuickMath Automatic Math Solutions (http://www.quickmath.com)
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.