$\displaystyle\lim_{x\to 0} \left [ \dfrac {1}{x^2} - \cot ^2 x \right ] = \lim_{x\to 0} \left [ \dfrac {1}{x^2} - \dfrac {1}{\tan ^2 x} \right ]$ $\displaystyle =\lim_{x\to 0} \left [ \dfrac {\tan^2 x-x^2}{x^2 \tan^2 x} \right ] \left [ \dfrac {0}{0} \right ]$ $\displaystyle = \lim_{x\to 0} \left [ \dfrac {\tan ^2 x - x^2}{x^4} \right ] \left ( \dfrac {x^2} {\tan ^2 x} \right ) \left [ \dfrac {0}{0} \right ]$ $\displaystyle=\lim_{x\to 0} \left [ \dfrac {\tan ^2 x- x^2}{x^4} \right ] \lim_{x\to 0} \left ( \dfrac {x^2}{\tan ^2 x} \right ) \left [ \dfrac {0}{0} \right ]$ $\displaystyle=\lim_{x\to 0} \left [ \dfrac {2 \tan x \sec^2 x - 2x}{4x^3} \right ] \lim_{x\to 0} \left ( \dfrac {2x}{2 \tan x \sec ^2 x} \right ) \left [ \dfrac {0}{0} \right ] $ $\displaystyle=\lim_{x\to 0} \left [ \dfrac {\sec^4 x + 2 \tan^2 x \sec^2 x-1}{6x^2} \right ] \left [ \dfrac {0}{0} \right ] \cdot \lim_{x\to 0} \left ( \dfrac {x}{\sec^4 x + 2 \tan^2 x \sec^2 x} \right )$ $\displaystyle=\lim_{x\to 0} \left [ \dfrac {4\sec^4 x \tan x + 4\tan x \sec^4 x + 4\tan^3 x \sec ^3 x}{12x} \right ] \left [ \dfrac {0}{0} \right ] \cdot (1)$ $\displaystyle=\lim_{x\to 0} \left [ \dfrac {2\tan x \sec^4 x + \tan^3 x \sec^2 x}{3x} \right ]$ $\displaystyle= \lim_{x\to 0} \left [ \dfrac {2 \tan x \sec^4 x}{3x} \right ] + \lim_{x\to 0} \left [ \dfrac{\tan x}{3} \right ] \tan^2 x \sec^2 x $ Applying limits. $=2/3+0$ $= 2/3$ Hence, $\displaystyle \lim_{x\to 0} \left [ \dfrac {1}{x^2} - \cot ^2 x \right ] = 2/3$