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
 A: I am not sure that this is an answer but, in any manner, it is too long for a comment.
Rewriting $$A=\frac{1}{\sin^2(x)}-\frac{1}{\sinh^2(x)}=\frac{\sinh^2(x)-\sin^2(x)}{\sin^2(x)\sinh^2(x)}$$ let us define $u=\sinh^2(x)-\sin^2(x)$ and $v=\sin^2(x)\sinh^2(x)$ (in order to use L'Hospital rule) and so let us compute the derivatives to be simplified one after eachother. So $$u=\sinh^2(x)-\sin^2(x)$$ $$\frac{du}{dx}=\sinh (2 x)-\sin (2 x)$$ $$\frac{d^2u}{dx^2}=2 (\cosh (2 x)- \cos (2 x))$$ $$\frac{d^3u}{dx^3}=4 (\sin (2 x)+\sinh (2 x))$$ $$\frac{d^4u}{dx^4}=8 (\cos (2 x)+\cosh (2 x))$$ At this point, we do not have any more $0$ as a result. So, we need to compute the fourth derivative of $v$ which is less pleasant but still doable $$v=\sin^2(x)\sinh^2(x)$$ $$\frac{dv}{dx}=2 \sin (x) \sinh (x) (\cos (x) \sinh (x)+\sin (x) \cosh (x))$$ $$\frac{d^2v}{dx^2}=-\cos (2 x)+\cosh (2 x)+2 \sin (2 x) \sinh (2 x)$$ $$\frac{d^3v}{dx^3}=2 (2 \cos (2 x)+1) \sinh (2 x)+2 \sin (2 x) (2 \cosh (2 x)+1)$$ $$\frac{d^4v}{dx^4}=4 (\cos (2 x)+(4 \cos (2 x)+1) \cosh (2 x))$$ So, after four steps of derivation $$L=\lim_{x \to 0} \Big(\frac{1}{\sin^2(x)}-\frac{1}{\sinh^2(x)}\Big)=\frac{16}{24}=\frac 2 3$$
Even if not using Taylor expansions, we knew a priori that we had to go up to the fourth derivative since, close to $x=0$, $\sin(x)\approx x$ and  $\sinh(x)\approx x$ making $v\approx x^4$.
A: 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*}
A: For every $x$ with $0<|x|<\pi$ we have
$$
\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{x-\sin x}{x^2\sin x}\cdot\frac{x+\sin x}{\sin x}
$$
Using l'Hopital's rule we have:
\begin{eqnarray}
\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}&=&\lim_{x\to0}\frac{(x-\sin x)'}{(x^2\sin x)'}=\lim_{x\to0}\frac{1-\cos x}{2x\sin x+x^2\cos x}=\lim_{x\to0}\frac{(1-\cos x)'}{(2x\sin x+x^2\cos x)'}\\
&=&\lim_{x\to0}\frac{\sin x}{2\sin x+4x\cos x-x^2\sin x}=\lim_{x\to0}\frac{(\sin x)'}{(2\sin x+4x\cos x-x^2\sin x)'}\\
&=&\lim_{x\to0}\frac{\cos x}{2\cos x+4\cos x-4x\sin x-2x\sin x-x^2\cos x}=\frac{1}{2+4}=\frac16\\
\lim_{x\to0}\frac{x+\sin x}{\sin x}&=&\lim_{x\to0}\left(\frac{x}{\sin x}+1\right)=\lim_{x\to0}\left[\frac{(x)'}{(\sin x)'}+1\right]=\left(\frac{1}{\cos x}+1\right)=\frac11+1=2.
\end{eqnarray}
It follows that
$$\tag{1}
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=\left(\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}\right)\cdot\left(\lim_{x\to0}\frac{x+\sin x}{\sin x}\right)=\frac16\cdot2=\frac13.
$$
Similarly, for every $x\ne 0$ we have
$$
\frac{1}{\sinh^2x}-\frac{1}{x^2}=\frac{x^2-\sinh^2x}{x^2\sinh^2x}=\frac{x-\sinh x}{x^2\sinh x}\cdot\frac{x+\sinh x}{\sinh x}
$$
Using l'Hopital's rule we have:
\begin{eqnarray}
\lim_{x\to0}\frac{x-\sinh x}{x^2\sinh x}&=&\lim_{x\to0}\frac{(x-\sinh x)'}{(x^2\sinh x)'}=\lim_{x\to0}\frac{1-\cosh x}{2x\sinh x+x^2\cosh x}=\lim_{x\to0}\frac{(1-\cosh x)'}{(2x\sinh x+x^2\cosh x)'}\\
&=&\lim_{x\to0}\frac{-\sinh x}{2\sinh x+4x\cosh x+x^2\sinh x}=\lim_{x\to0}\frac{(\sinh x)'}{(2\sinh x+4x\cosh x-x^2\sinh x)'}\\
&=&\lim_{x\to0}\frac{-\cosh x}{2\cosh x+4\cosh x+4x\sinh x+2x\sinh x+x^2\cosh x}=\frac{-1}{2+4}=-\frac16\\
\lim_{x\to0}\frac{x+\sinh x}{\sinh x}&=&\lim_{x\to0}\left(\frac{x}{\sinh x}+1\right)=\lim_{x\to0}\left[\frac{(x)'}{(\sinh x)'}+1\right]=\left(\frac{1}{\cosh x}+1\right)=\frac11+1=2.
\end{eqnarray}
It follows that
$$\tag{2}
\lim_{x\to0}\left(\frac{1}{\sinh^2x}-\frac{1}{x^2}\right)=\left(\lim_{x\to0}\frac{x-\sinh x}{x^2\sinh x}\right)\cdot\left(\lim_{x\to0}\frac{x+\sinh x}{\sinh x}\right)=-\frac16\cdot2=-\frac13.
$$
Combining (1) and (2) we get:
\begin{eqnarray}
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{\sinh^2x}\right)&=&\lim_{x\to0}\left[\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)-\left(\frac{1}{\sinh^2x}-\frac{1}{x^2}\right)\right]\\
&=&\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)-\lim_{x\to0}\left(\frac{1}{\sinh^2x}-\frac{1}{x^2}\right)\\
&=&\frac13-\left(-\frac13\right)=\frac13+\frac13=\frac23.
\end{eqnarray}
A: Your error is in the last line when you think that $\lim_{x\to0}\frac{x+\sin x}{x\sin x}=0$. Since $\sin x\sim x$ for small $x$, we have in fact $\frac{x+\sin x}{x\sin x}\sim\frac{x+x}{x^2}=\frac 2x\not \to 0$
