Lim x→0((1)/(x^2-cot ^2x) is equal to

Use $\cot^2x=\dfrac{\cos^2x}{\sin^2x}$, so $\dfrac1{x^2}-\cot^2x=\dfrac{\sin^2x-x^2\cos^2x}{x^2\sin^2x}$. With the series $\sin^2x=x^2-\dfrac{x^4}{3}+\cdots$ and $x^2\cos^2x=x^2-x^4+\cdots$ the numerator is $\dfrac23x^4$ and the denominator is $x^4$ as $x\to0$, hence the limit is $\dfrac23$. The value $\dfrac13$ is what you get by dropping the $\cos^2x$ expansion's $x^4$ term, and $1$ treats the two terms as cancelling at the leading order only.