# lim_(x->0)(1/x^2-1/sin^2x)=  ?

##### 1 Answer
Nov 26, 2016

$- \frac{1}{3}$

#### Explanation:

We have

$\frac{1}{x} ^ 2 - \frac{1}{\sin} ^ 2 x = \frac{{\sin}^{2} x - {x}^{2}}{{x}^{2} {\sin}^{2} x} = \frac{{\left(\sin \frac{x}{x}\right)}^{2} - 1}{1 - {\cos}^{2} x} =$
$= \left(\frac{\sin \frac{x}{x} + 1}{1 + \cos x}\right) \left(\frac{\sin \frac{x}{x} - 1}{1 - \cos x}\right)$

Now following with ${\lim}_{x \to 0} \left(\frac{\sin \frac{x}{x} - 1}{1 - \cos x}\right)$

At this point we introduce the series representation for $\sin x$ and $\cos x$

sinx = x-x^3/(3!)+x^5/(5!)+cdots
cosx = 1-x^2/(2!)+x^4/(4!)+cdots

so

sinx/x = 1-x^2/(3!)+x^4/(5!)+cdots

making substitutions

(sinx/x-1)/(1-cosx)=(-x^2/(!3)+x^4/(5!)-x^6/(7!)+cdots)/(x^2/(2!)-x^4/(4!)+x^6/(6!)+cdots)=(-1/(!3)+x^2/(5!)-x^4/(7!)+cdots)/(1/(2!)-x^2/(4!)+x^4/(6!)+cdots)

then

${\lim}_{x \to 0} \frac{\sin \frac{x}{x} - 1}{1 - \cos x} = - \frac{1}{3}$ and
${\lim}_{x \to 0} \frac{\sin \frac{x}{x} + 1}{1 + \cos x} = 1$

Finally

${\lim}_{x \to 0} \left(\frac{\sin \frac{x}{x} + 1}{1 + \cos x}\right) \left(\frac{\sin \frac{x}{x} - 1}{1 - \cos x}\right) = \left(1\right) \left(- \frac{1}{3}\right) = - \frac{1}{3}$