clara1223
  • clara1223
find the limit as x approaches 0 of (2x^2)/(tan^2)(9x)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
let me guess, no l'hopital?
clara1223
  • clara1223
not yet
thomas5267
  • thomas5267
\[ \lim_{x\to0}\frac{2x^2}{\tan^2(9x)}? \]

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clara1223
  • clara1223
yes
anonymous
  • anonymous
does it help to know that \[\lim_{x\to 0}\frac{x}{\tan(x)}=1\]?
anonymous
  • anonymous
by which i really mean, can you use that?
clara1223
  • clara1223
Yes we can use that, I'm just not sure how to get it to look like that
anonymous
  • anonymous
algea bra pull out the 2, make a product out of it
clara1223
  • clara1223
could I also pull out the 9?
clara1223
  • clara1223
to make a product of 2/9?
anonymous
  • anonymous
no the 9 is inside the tangent, but you can multiply top and bottom by 9
anonymous
  • anonymous
or maybe you need 81 since there are two of them
anonymous
  • anonymous
\[\frac{2}{81}\frac{9x}{\tan(9x)}\times \frac{9x}{\tan(9x)}\]
clara1223
  • clara1223
so the answer is 2/81 x 1 x 1?
anonymous
  • anonymous
yup
thomas5267
  • thomas5267
\[ \begin{align*} &\phantom{{}={}}\lim_{x\to0}\frac{2x^2}{\tan^2(9x)}\\ &=2\left(\lim_{x\to0}\frac{x}{\tan(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x\cos(9x)}{\sin(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x}{\sin(9x)}\right)^2\left(\lim_{x\to0}\cos(9x)\right)^2\\ &=2\left(\lim_{x\to0}\frac{1}{\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\frac{1}{\lim_{x\to0}\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\lim_{x\to0}\frac{\sin(9x)}{x}\right)^{-2}\\ &=2\left(\lim_{x\to0}\frac{9\sin(9x)}{9x}\right)^{-2}\\ &=2\left(9\lim_{x\to0}\frac{\sin(9x)}{9x}\right)^{-2}\\ &=2\left(9\right)^{-2}\\ &=\frac{2}{81} \end{align*} \]

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