<s>[INST] Limit Infinity Minus Infinity form without using infinite series Expansions Evaluate $$L=\lim_{x \to 0} \frac{1}{\sin^2x}-\frac{1}{\sinh^2x}$$ If we take L.C.M and use LHopital's Rule it becomes quite Tedious. I tried in this way.
$$
\sinh x=\frac{e^x-e^{-x}}{2}
$$ 
Now 
$$\lim_{x \to 0}\frac{e^x-e^{-x}}{x}=2 \implies \lim_{x \to 0}\left(\frac{e^x-e^{-x}}{x}\right)^{-2}=\frac{1}{4}
$$
Now 
$$
\sinh x=\left(\frac{e^x-e^{-x}}{x}\right)\left(\frac{x}{2}\right)
$$ 
So
$$
\lim_{x \to 0}\frac{1}{\sinh^2x}=\lim_{x \to 0}\left(\frac{e^x-e^{-x}}{x}\right)^{-2}\:\lim_{x \to 0}\frac{4}{x^2}=\lim_{x \to 0}\frac{1}{x^2}
$$ 
Hence
$$
L=\lim_{x \to 0}\frac{1}{\sin^2x}-\frac{1}{x^2}=\lim_{x \to 0}\frac{(x-\sin x)(x+\sin x)}{(x\sin x)(x\sin x)}=\lim_{x \to 0}\frac{x-\sin x}{x\sin x}\:\lim_{x \to 0}\frac{x+\sin x}{x\sin x}$$
Individually if i evaluate above limits I am getting Zero. But answer is not correct. Can anyone spot out where I am going wrong
 [/INST] Write the original expression as
\begin{eqnarray*}
\frac{1}{\sin ^{2}x}-\frac{1}{\sinh ^{2}x} &=&\frac{\sinh ^{2}x-\sin ^{2}x}{%
\sin ^{2}x\sinh ^{2}x} \\
&=&\left( \frac{\sinh x-\sin x}{x^{3}}\right) \times \left( \frac{\sinh
x+\sin x}{x}\right) \times \left( \frac{x}{\sin x}\right) ^{2}\times \left( 
\frac{x}{\sinh x}\right) ^{2}.
\end{eqnarray*}
Using L'Hospital rules successively one obtains
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sinh x-\sin x}{x^{3}}\right)
=\lim_{x\rightarrow 0}\left( \frac{\cosh x-\cos x}{3x^{2}}\right)
=\lim_{x\rightarrow 0}\left( \frac{\sinh x+\sin x}{6x}\right)
=\lim_{x\rightarrow 0}\left( \frac{\cosh x+\cos x}{6}\right) =\frac{1+1}{6}=%
\frac{1}{3}
\end{equation*}
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sinh x+\sin x}{x}\right)
=\lim_{x\rightarrow 0}\left( \frac{\cosh x+\cos x}{1}\right) =\frac{1+1}{1}=2
\end{equation*}
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{x}{\sin x}\right) ^{2}=\left(
\lim_{x\rightarrow 0}\frac{x}{\sin x}\right) ^{2}=\left( \lim_{x\rightarrow
0}\frac{1}{\cos x}\right) ^{2}=\left( 1\right) ^{2}=1
\end{equation*}
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{x}{\sinh x}\right) ^{2}=\left(
\lim_{x\rightarrow 0}\frac{x}{\sinh x}\right) ^{2}=\left( \lim_{x\rightarrow
0}\frac{1}{\cosh x}\right) ^{2}=\left( 1\right) ^{2}=1
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{1}{\sin ^{2}x}-\frac{1}{\sinh ^{2}x}%
\right) =\left( \frac{1}{3}\right) \times \left( 2\right) \times \left(
1\right) \times \left( 1\right) =\frac{2}{3}
\end{equation*}
</s>