Find the limit of this trigonometric expression as x approaches 0 $$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})(\cot(x)+\frac{1}{x})$$
I have computed the limits
$$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})=0$$
$$\lim_{x\to 0^+} (\cot(x)+\frac{1}{x})=\infty$$
I can't multiply the limits together.
I rewrote the initial expression:
$$\lim_{x\to 0^+} (\frac{x\cos(x)-\sin(x)}{x\sin(x)})(\frac{x\cos(x)+\sin(x)}{x\sin(x)})$$
$$\lim_{x\to 0^+} \frac{x^{2}\cos^{2}(x)-\sin^{2}(x)}{x^{2}\sin^{2}(x)}$$
$$\lim_{x\to 0^+} \frac{\cos^{2}(x)-1}{\sin^{2}(x)}$$
Using Pythagorean Identity
$$\lim_{x\to 0^+} \frac{-\sin^{2}(x)}{\sin^{2}(x)}=-1$$
I also got the same result using L'Hospital's Rule.
I know the limit is $\frac{-2}{3}$ 
What did I do wrong?
 A: A simple way to do it is based on Taylor series; since $$\cot(x)=\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ $$\cot(x)-\frac{1}{x}=-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ $$\cot(x)+\frac{1}{x}=\frac{2}{x}-\frac{x}{3}-\frac{x^3}{45}+O\left(x^4\right)$$ Making the product $$(\cot(x)-\frac{1}{x})\,(\cot(x)+\frac{1}{x})=-\frac{2}{3}+\frac{x^2}{15}+O\left(x^3\right)$$ which shows the limit and how it is approached.
A: It is possible to do not use Taylor series. First write
\begin{eqnarray*}
\cot ^{2}x-\frac{1}{x^{2}} &=&\left( \frac{x\cos x-\sin x}{x\sin x}\right)
\left( \frac{x\cos x+\sin x}{x\sin x}\right)  \\
&=&\frac{x^{2}}{\sin ^{2}x}\left( \frac{x\cos x-\sin x}{x^{3}}\right) \left( 
\frac{x\cos x+\sin x}{x}\right)  \\
&=&\left( \frac{x}{\sin x}\right) ^{2}\left( \frac{x\left( \cos x-1\right)
+(x-\sin x)}{x^{3}}\right) \left( \cos x+\frac{\sin x}{x}\right)  \\
&=&\left( \frac{x}{\sin x}\right) ^{2}\left( \frac{\cos x-1}{x^{2}}+\frac{%
x-\sin x}{x^{3}}\right) \left( \cos x+\frac{\sin x}{x}\right) 
\end{eqnarray*}
Basic limits as
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{x}{\sin x} &=&1 \\
\
lim_{x\rightarrow 0}\frac{\cos x-1}{x^{2}} &=&-\frac{1}{2} \\
\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}} &=&\frac{1}{6} \\
\lim_{x\rightarrow 0}\cos x &=&1
\end{eqnarray*}
imply that
\begin{eqnarray*}
\lim_{x\rightarrow 0}\left( \cot ^{2}x-\frac{1}{x^{2}}\right)  &=&\left(
\lim_{x\rightarrow 0}\frac{x}{\sin x}\right) ^{2}\left( \lim_{x\rightarrow 0}%
\frac{\cos x-1}{x^{2}}+\lim_{x\rightarrow 0}\frac{x-\sin x}{x^{3}}\right)
\left( \lim_{x\rightarrow 0}\cos x+\lim_{x\rightarrow 0}\frac{\sin x}{x}%
\right)  \\
&=&\left( 1\right) ^{2}\left( -\frac{1}{2}+\frac{1}{6}\right) \left(
1+1\right) =-\frac{2}{3}.
\end{eqnarray*}
A: since  $\sin x$ has a very familiar Maclaurin expansion one may note:
$$
\cot^2 x -\frac1{x^2}= (\sin x)^{-2} - 1 -\frac1{x^2}= x^{-2}(1-\frac{x^2}6+\dots)^{-2} -1 -\frac1{x^2} \\
=x^{-2}(1 +(-2)(-\frac{x^2}6)+O(x^4))) - 1-\frac1{x^2}  \\
=-\frac23+O(x^2)
$$
