## anonymous one year ago find the limit as x approaches 0 of (2x^2)/(tan^2)(9x)

1. anonymous

let me guess, no l'hopital?

2. anonymous

not yet

3. thomas5267

$\lim_{x\to0}\frac{2x^2}{\tan^2(9x)}?$

4. anonymous

yes

5. anonymous

does it help to know that $\lim_{x\to 0}\frac{x}{\tan(x)}=1$?

6. anonymous

by which i really mean, can you use that?

7. anonymous

Yes we can use that, I'm just not sure how to get it to look like that

8. anonymous

algea bra pull out the 2, make a product out of it

9. anonymous

could I also pull out the 9?

10. anonymous

to make a product of 2/9?

11. anonymous

no the 9 is inside the tangent, but you can multiply top and bottom by 9

12. anonymous

or maybe you need 81 since there are two of them

13. anonymous

$\frac{2}{81}\frac{9x}{\tan(9x)}\times \frac{9x}{\tan(9x)}$

14. anonymous

so the answer is 2/81 x 1 x 1?

15. anonymous

yup

16. thomas5267

\begin{align*} &\phantom{{}={}}\lim_{x\to0}\frac{2x^2}{\tan^2(9x)}\\ &=2\left(\lim_{x\to0}\frac{x}{\tan(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x\cos(9x)}{\sin(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x}{\sin(9x)}\right)^2\left(\lim_{x\to0}\cos(9x)\right)^2\\ &=2\left(\lim_{x\to0}\frac{1}{\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\frac{1}{\lim_{x\to0}\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\lim_{x\to0}\frac{\sin(9x)}{x}\right)^{-2}\\ &=2\left(\lim_{x\to0}\frac{9\sin(9x)}{9x}\right)^{-2}\\ &=2\left(9\lim_{x\to0}\frac{\sin(9x)}{9x}\right)^{-2}\\ &=2\left(9\right)^{-2}\\ &=\frac{2}{81} \end{align*}