##\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 $\lim_{ x \to 0}\left(\dfrac {1}{\sin^2x}-\dfrac1{x^2}\right)=L$

Let $x=2y$

$\lim_{ y \to 0}\left(\dfrac {1}{\sin^22y}-\dfrac1{4y^2}\right)=\lim_{ y \to 0}\left(\dfrac1{4\sin^2y\cdot\cos^2y}-\dfrac1{4y^2}\right)=L$

$=\dfrac14\lim_{ y \to 0}\left(\dfrac1{\cos^2{y}}+\dfrac1{\sin^2{y}}-\dfrac1{y^2}\right)=\dfrac14+\dfrac{L}{4}=L$

$L=\dfrac13$

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 $1/3$. 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.
$\lim_{x \to0} \left(\dfrac{1}{\sin^2 x}-\dfrac{1}{x^2}\right)=\lim_{x \to0}\dfrac{x(x-\sin x)(x+\sin x)}{x^3\sin^2x}$

$\lim_{x \to0}\left(\dfrac{x}{\sin x}\right)\cdot\lim_{x \to0}\dfrac{x+\sin x}{\sin x}\cdot\lim_{x \to0}\dfrac{1-\cos x}{3x^2}$

$=1.\lim_{x \to0}\dfrac{1+\cos x}{\cos x}\cdot\lim_{x \to0}\dfrac{1-\cos x}{3x^2}=1.2.\lim_{x \to0}\dfrac{\sin x}{6x}=\dfrac 13$

 

[PeroK]

The clue was to look at the Taylor series for $\sin x$:
$$\sin x = x - \frac{x^3}{6} + \dots$$Hence:$$x - \sin x = \frac{x^3}{6} + \dots$$And:
$$\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac 1 6$$Which you can confirm formally using L'Hopital.