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anonymous
 one year ago
calculus exam tomorrow
please help me,
calculate :
anonymous
 one year ago
calculus exam tomorrow please help me, calculate :

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0}[1/\sin ^{2}x1/x^{2}]\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Using the fact that \(\lim\limits_{x\to0}\dfrac{\sin x}{x}=1\), you have that \(\sin x\approx x\) near \(x=0\), so \[\lim_{x\to0}\left(\frac{1}{\sin^2x}\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{1}{x^2}\frac{1}{x^2}\right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles the answer would be 0? but on the answer sheet is 1/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm, 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles am doing like this too.. and the derivative is driving me crazy.. lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{\dfrac{d}{dx}\left[x^2\sin^2x\right]}{\dfrac{d}{dx}\left[x^2\sin^2x\right]}=\frac{2x2\sin x\cos x}{2x\sin^2x+2x^2\sin x\cos x}=\frac{2x\sin2x}{2x\sin^2x+x^2\sin2x}\] 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: \[\begin{align*}\frac{\dfrac{d}{dx}\left[2x\sin2x\right]}{\dfrac{d}{dx}\left[2x\sin^2x+x^2\sin2x\right]}&=\frac{22\cos2x}{2\sin^2x+4x\sin x\cos x+2x\sin2x+2x^2\cos2x}\\\\ &=\frac{22\cos2x}{2\sin^2x+4x\sin2x+2x^2\cos2x} \end{align*}\] and keep this up.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles the final answer is correct 1/3, but i wonder why the first solution is incorrect .

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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\dfrac{1}{3}\), not \(\dfrac{1}{2}\)*

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yup, 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 indepth discussion: http://math.stackexchange.com/a/105606/170231

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In 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, \[\begin{align*}\frac{1}{\sin ^2x}\frac{1}{x^2}&\approx\frac{1}{x^2\dfrac{1}{3}x^4}\frac{1}{x^2}\\\\ &=\frac{x^2x^2+\dfrac{1}{3}x^4}{x^2\left(x^2\dfrac{1}{3}x^4\right)}\\\\ &=\frac{1}{3}\times\frac{1}{1\dfrac{1}{3}x^2}\\\\ &\to\frac{1}{3}\text{ as }x\to0 \end{align*}\]
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