$$\lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-cot^2x \Bigg] \;=\; \lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-\dfrac{cos^2}{sin^2x} \Bigg] $$ $$\lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-cot^2x \Bigg] \;=\; \lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-\dfrac{1-sin^2x}{sin^2x} \Bigg] $$ $$ = \; \lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-\dfrac{1}{sin^2x} + 1 \Bigg] $$ $$ = \; \lim_{x\to 0} \Bigg[ \dfrac{sin^2x - x^2}{x^2 sin^2x} + \lim_{x\to 0}(1) \Bigg] $$ Applying L’ hospital’s Rule, $$ \therefore \lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-cot^2x \Bigg] \;=\; \lim_{x\to 0} \Bigg[ \dfrac{2sinxcosx - 2x}{ x^2(2sinxcosx) +2xsin^2x} \Bigg] + 1 $$ $$ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{sin2x - 2x}{ x^2 sin2x +2xsin^2x} \Bigg] + 1 $$ Applying L’ Hospital’s rule again, $$ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{2cos2x - 2}{ 2x^2 cos2x + 2xsin2x + 4xsinx.cosx +2sin^2x} \Bigg] + 1 $$ $$ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{2cos2x - 2}{ 2x^2 cos2x + 2xsin2x + 2xsin2x +2sin^2x} \Bigg] + 1 $$ $$ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{2cos2x - 2}{ 2x^2 cos2x + 4xsin2x + +2sin^2x} \Bigg] + 1 $$ $$ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{cos2x - 1}{ x^2 cos2x + 2xsin2x + +sin^2x} \Bigg] + 1 $$ Applying L’ hospital’s Rule, $$ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{ -2sin2x }{ -2x^2 sin2x + 2xcos2x + 4xcos2x + 2sin2x +2sinx \;cosx} \Bigg] + 1 \ \; \ \; \ \;=\; \lim_{x\to 0} \Bigg[ \dfrac{ -2sin2x }{ -2x^2 sin2x + 6xcos2x + 2sin2x + sin2x} \Bigg] + 1 \ \; \ \; \
\;=\; \lim_{x\to 0} \Bigg[ \dfrac{ -2sin2x }{ -2x^2 sin2x + 6xcos2x + 3sin2x } \Bigg] + 1 $$ Applying L’ Hospital’s Rule, $$ \lim_{x\to 0} \Bigg[ \dfrac{ -4cos2x }{ -4x^2 cos2x -4xcos2x - 12xcos2x + 6cos2x + 6cos2x } \Bigg] + 1 $$

As $ x \rightarrow 0 , x \ne 0 $ Substituting x=0 in the limit,

$$ \therefore \lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-cot^2x \Bigg] \;=\; \dfrac{-4(1)}{ 0-0-0+6(1)+6 } + 1 $$ $$ = \dfrac{-4}{12} + 1 \;=\; 1- \dfrac{1}{3} \;=\; \dfrac{2}{3} $$ $$ \therefore \lim_{x\to 0} \Bigg[ \dfrac{1}{x^2}-cot^2x \Bigg] \;=\; \dfrac{2}{3} $$