Question

calculus exam tomorrow
please help me,
calculate :

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

\(\dfrac{1}{3}\), not \(\dfrac{1}{2}\)*

Answer 10

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

Answer 11

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,
\[\begin{align*}\frac{1}{\sin ^2x}-\frac{1}{x^2}&\approx\frac{1}{x^2-\dfrac{1}{3}x^4}-\frac{1}{x^2}\\\\
&=\frac{x^2-x^2+\dfrac{1}{3}x^4}{x^2\left(x^2-\dfrac{1}{3}x^4\right)}\\\\
&=\frac{1}{3}\times\frac{1}{1-\dfrac{1}{3}x^2}\\\\
&\to\frac{1}{3}\text{ as }x\to0
\end{align*}\]