$\lim \limits_{x \to 0} \frac{1}{x^2}-\csc^2x $ using L'Hôpital's rule

Question

This is from an exercise set on L'Hôpital's rule. I've never seen anything so stubborn and recalcitrant. Anybody have suggestions?

$\lim \limits_{x \to 0} \frac{1}{x^2}-\csc^2x $

Answer 1

HINT: Write this as $$\lim_{x\to 0}\left(\frac1{x^2} - \frac1{\sin^2x}\right)$$ and find a common denominator. Further hint: Using Taylor polynomials is way better than using L'Hôpital's rule 4 times.

Answer 2

First, write \begin{equation*} \frac{1}{x^{2}}-\frac{1}{\sin ^{2}x}=\frac{\sin ^{2}x-x^{2}}{x^{2}\sin ^{2}x}% =\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\sin x+x}{x}\right) \left( \frac{x}{\sin x}\right) ^{2} \end{equation*} Next evaluate the following limits using L'Hospital's rule \begin{equation*} \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}=\lim_{x\rightarrow 0}\frac{\cos x-1}{3x^{2}}=\lim_{x\rightarrow 0}\frac{-\sin x}{6x}=\lim_{x\rightarrow 0} \frac{-\cos x}{6}=\frac{-1}{6} \end{equation*} \begin{equation*} \lim_{x\rightarrow 0}\frac{x}{\sin x}=\lim_{x\rightarrow 0}\frac{1}{\cos x}=1 \end{equation*} Next \begin{eqnarray*} \lim_{x\rightarrow 0}\frac{1}{x^{2}}-\frac{1}{\sin ^{2}x} &=&\lim_{x% \rightarrow 0}\left( \frac{\sin x-x}{x^{3}}\right) \left( \frac{\sin x+x}{x} \right) \left( \frac{x}{\sin x}\right) ^{2} \\ &=&\left( \lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}}\right) \left( \lim_{x\rightarrow 0}\frac{\sin x}{x}+1\right) \left( \lim_{x\rightarrow 0} \frac{x}{\sin x}\right) ^{2} \\ &=&\left( \frac{-1}{6}\right) \left( 1+1\right) \left( 1\right) ^{2}=-\frac{1 }{3}. \end{eqnarray*}

Answer 3

As $x\to0$,

$$\frac{1}{x^2}-\frac{1}{\sin^2 x}=\frac{1}{x^2}-\frac{1}{(x-x^3/6+o(x^3))^2}=\frac{1}{x^2}-\frac{1}{x^2-x^4/3+o(x^4)}\\=\frac{1}{x^2}-\frac{1}{x^2}\left(\frac{1}{1-x^2/3+o(x^2)}\right)=\frac{1}{x^2}-\frac{1}{x^2}\left(1+x^2/3+o(x^2)\right)=-\frac{1}{3}+o(1)\to-\frac{1}{3}$$