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clara1223
 one year ago
find the limit as x approaches 0 of (2x^2)/(tan^2)(9x)
clara1223
 one year ago
find the limit as x approaches 0 of (2x^2)/(tan^2)(9x)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let me guess, no l'hopital?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \lim_{x\to0}\frac{2x^2}{\tan^2(9x)}? \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does it help to know that \[\lim_{x\to 0}\frac{x}{\tan(x)}=1\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0by which i really mean, can you use that?

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0Yes we can use that, I'm just not sure how to get it to look like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0algea bra pull out the 2, make a product out of it

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0could I also pull out the 9?

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0to make a product of 2/9?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no the 9 is inside the tangent, but you can multiply top and bottom by 9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or maybe you need 81 since there are two of them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{2}{81}\frac{9x}{\tan(9x)}\times \frac{9x}{\tan(9x)}\]

clara1223
 one year ago
Best ResponseYou've already chosen the best response.0so the answer is 2/81 x 1 x 1?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \begin{align*} &\phantom{{}={}}\lim_{x\to0}\frac{2x^2}{\tan^2(9x)}\\ &=2\left(\lim_{x\to0}\frac{x}{\tan(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x\cos(9x)}{\sin(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x}{\sin(9x)}\right)^2\left(\lim_{x\to0}\cos(9x)\right)^2\\ &=2\left(\lim_{x\to0}\frac{1}{\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\frac{1}{\lim_{x\to0}\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\lim_{x\to0}\frac{\sin(9x)}{x}\right)^{2}\\ &=2\left(\lim_{x\to0}\frac{9\sin(9x)}{9x}\right)^{2}\\ &=2\left(9\lim_{x\to0}\frac{\sin(9x)}{9x}\right)^{2}\\ &=2\left(9\right)^{2}\\ &=\frac{2}{81} \end{align*} \]
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