##\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)##

[littlemathquark]

Homework Statement

Find $\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)$

Relevant Equations

$\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)$

My solution is:

Let $$

Let $$

$$

$$

$$

But I think I must show that firstly this limit L is a infinity number but I don't know how. Please help.

 

[PeroK]

You've shown that if the limit is finite, then it is $$. It looks tricky to prove that the limit is finite.

You could use l'Hopital's rule several times. Although perhaps there is something better than that.

 

[littlemathquark]

My second solution using L'Hopital rule twice.
$$

$$

$$

 

[PeroK]

The clue was to look at the Taylor series for $$:
$$Hence:$$And:
$$Which you can confirm formally using L'Hopital.

 

[pasmith]

For sufficiently small x^2,
\begin{split}<br /> \frac{1}{\sin^2 x} - \frac 1{x^2} &amp;= (\sin x)^{-2} - \frac{1}{x^2} \\<br /> &amp;= \frac{1}{x^2} \left(1 - \frac{x^2}{6} + O(x^4)\right)^{-2} - \frac{1}{x^2} \\<br /> &amp;= \frac{1}{x^2} \left( 1 + \frac{x^2}{3} + O(x^4)\right) - \frac{1}{x^2} \\<br /> &amp;= \frac{1}{3} + O(x^2) \end{split}

 

[pasmith]

Another method is to set f(x) = (\sin x)/x with f(0) = 1. Then <br /> \frac{1}{\sin^2 x} - \frac{1}{x^2} = \frac{1}{x^2f^2} - \frac1{x^2} = \frac{1}{f^2} \frac{1 - f^2}{x^2} and <br /> f(x) = 1 - \frac{x^2}{6} + \dots \Rightarrow 1 - f^2(x) = \frac{x^2}{3} + \dots