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

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1. anonymous

$\lim_{x \rightarrow 0}[1/\sin ^{2}x-1/x^{2}]$

2. anonymous

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

@SithsAndGiggles the answer would be 0? but on the answer sheet is 1/3

4. anonymous

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

@SithsAndGiggles am doing like this too.. and the derivative is driving me crazy.. lol

6. anonymous

$\frac{\dfrac{d}{dx}\left[x^2-\sin^2x\right]}{\dfrac{d}{dx}\left[x^2\sin^2x\right]}=\frac{2x-2\sin x\cos x}{2x\sin^2x+2x^2\sin x\cos x}=\frac{2x-\sin2x}{2x\sin^2x+x^2\sin2x}$ 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

@SithsAndGiggles the final answer is correct 1/3, but i wonder why the first solution is incorrect .

8. anonymous

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

$$\dfrac{1}{3}$$, not $$\dfrac{1}{2}$$*

10. anonymous

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

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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