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clara1223

  • one year ago

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

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  1. anonymous
    • one year ago
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    let me guess, no l'hopital?

  2. clara1223
    • one year ago
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    not yet

  3. thomas5267
    • one year ago
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    \[ \lim_{x\to0}\frac{2x^2}{\tan^2(9x)}? \]

  4. clara1223
    • one year ago
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    yes

  5. anonymous
    • one year ago
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    does it help to know that \[\lim_{x\to 0}\frac{x}{\tan(x)}=1\]?

  6. anonymous
    • one year ago
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    by which i really mean, can you use that?

  7. clara1223
    • one year ago
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    Yes we can use that, I'm just not sure how to get it to look like that

  8. anonymous
    • one year ago
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    algea bra pull out the 2, make a product out of it

  9. clara1223
    • one year ago
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    could I also pull out the 9?

  10. clara1223
    • one year ago
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    to make a product of 2/9?

  11. anonymous
    • one year ago
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    no the 9 is inside the tangent, but you can multiply top and bottom by 9

  12. anonymous
    • one year ago
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    or maybe you need 81 since there are two of them

  13. anonymous
    • one year ago
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    \[\frac{2}{81}\frac{9x}{\tan(9x)}\times \frac{9x}{\tan(9x)}\]

  14. clara1223
    • one year ago
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    so the answer is 2/81 x 1 x 1?

  15. anonymous
    • one year ago
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    yup

  16. thomas5267
    • one year ago
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    \[ \begin{align*} &\phantom{{}={}}\lim_{x\to0}\frac{2x^2}{\tan^2(9x)}\\ &=2\left(\lim_{x\to0}\frac{x}{\tan(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x\cos(9x)}{\sin(9x)}\right)^2\\ &=2\left(\lim_{x\to0}\frac{x}{\sin(9x)}\right)^2\left(\lim_{x\to0}\cos(9x)\right)^2\\ &=2\left(\lim_{x\to0}\frac{1}{\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\frac{1}{\lim_{x\to0}\frac{\sin(9x)}{x}}\right)^2\\ &=2\left(\lim_{x\to0}\frac{\sin(9x)}{x}\right)^{-2}\\ &=2\left(\lim_{x\to0}\frac{9\sin(9x)}{9x}\right)^{-2}\\ &=2\left(9\lim_{x\to0}\frac{\sin(9x)}{9x}\right)^{-2}\\ &=2\left(9\right)^{-2}\\ &=\frac{2}{81} \end{align*} \]

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