Question

find the limit as x approaches 0 of (2x^2)/(tan^2)(9x)

Answer 1

let me guess, no l'hopital?

Answer 2

not yet

Answer 3

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

Answer 4

yes

Answer 5

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

Answer 6

by which i really mean, can you use that?

Answer 7

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

Answer 8

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

Answer 9

could I also pull out the 9?

Answer 10

to make a product of 2/9?

Answer 11

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

Answer 12

or maybe you need 81 since there are two of them

Answer 13

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

Answer 14

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

Answer 15

yup

Answer 16

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