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anonymous

  • one year ago

calculus exam tomorrow please help me, calculate :

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  1. anonymous
    • one year ago
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    \[\lim_{x \rightarrow 0}[1/\sin ^{2}x-1/x^{2}]\]

  2. anonymous
    • one year ago
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    Using the fact that \(\lim\limits_{x\to0}\dfrac{\sin x}{x}=1\), you have that \(\sin x\approx x\) near \(x=0\), so \[\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{1}{x^2}-\frac{1}{x^2}\right)\]

  3. anonymous
    • one year ago
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    @SithsAndGiggles the answer would be 0? but on the answer sheet is 1/3

  4. anonymous
    • one year ago
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    Hmm, interesting... Let's try something else then. \[\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x}\to\frac{0}{0}\quad\text{as }x\to0\] Use L'Hopital's rule from here.

  5. anonymous
    • one year ago
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    @SithsAndGiggles am doing like this too.. and the derivative is driving me crazy.. lol

  6. anonymous
    • one year ago
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    \[\frac{\dfrac{d}{dx}\left[x^2-\sin^2x\right]}{\dfrac{d}{dx}\left[x^2\sin^2x\right]}=\frac{2x-2\sin x\cos x}{2x\sin^2x+2x^2\sin x\cos x}=\frac{2x-\sin2x}{2x\sin^2x+x^2\sin2x}\] Writing \(2\sin x\cos x=\sin2x\) might make things a bit simpler. We get another indeterminate form \(\dfrac{0}{0}\), so do it again: \[\begin{align*}\frac{\dfrac{d}{dx}\left[2x-\sin2x\right]}{\dfrac{d}{dx}\left[2x\sin^2x+x^2\sin2x\right]}&=\frac{2-2\cos2x}{2\sin^2x+4x\sin x\cos x+2x\sin2x+2x^2\cos2x}\\\\ &=\frac{2-2\cos2x}{2\sin^2x+4x\sin2x+2x^2\cos2x} \end{align*}\] and keep this up.

  7. anonymous
    • one year ago
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    @SithsAndGiggles the final answer is correct 1/3, but i wonder why the first solution is incorrect .

  8. anonymous
    • one year ago
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    I'm currently looking into it. It's an interesting pattern here, if you can call it that: \[f(x,n)=\frac{1}{\sin^nx}-\frac{1}{x^n}\] \[\begin{array}{c|c}n&\lim\limits_{x\to0}f(x,n)\\ \hline 0&0\\ 1&0\\ 2&\dfrac{1}{2}\\ \ge3&\infty \end{array}\] I suspect it has something to do with the approximation \(\sin x\approx x\) not being valid in some cases.

  9. anonymous
    • one year ago
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    \(\dfrac{1}{3}\), not \(\dfrac{1}{2}\)*

  10. anonymous
    • one year ago
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    Yup, I was on the right track. It has to do with the fact that we're neglecting to take into consideration the other terms in the approximation of \(\sin x\). See here for an in-depth discussion: http://math.stackexchange.com/a/105606/170231

  11. anonymous
    • one year ago
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    In practice: \[\sin x\approx x-\frac{1}{3!}x^3~~\implies~~\sin^2x\approx x^2-\frac{2}{3!}x^4=x^2-\frac{1}{3}x^4\] So, \[\begin{align*}\frac{1}{\sin ^2x}-\frac{1}{x^2}&\approx\frac{1}{x^2-\dfrac{1}{3}x^4}-\frac{1}{x^2}\\\\ &=\frac{x^2-x^2+\dfrac{1}{3}x^4}{x^2\left(x^2-\dfrac{1}{3}x^4\right)}\\\\ &=\frac{1}{3}\times\frac{1}{1-\dfrac{1}{3}x^2}\\\\ &\to\frac{1}{3}\text{ as }x\to0 \end{align*}\]

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