Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$ Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
 A: Well, you have 
$$\left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right) = \left(\frac{1}{x} - \frac{1}{\sin x}\right)\left(\frac{1}{x} + \frac{1}{\sin x}\right)$$
$$=\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)\left(1+ \frac{x}{\sin x}\right)$$
Since the limit of $\left(1+ \frac{x}{\sin x}\right)$ is $2$, your answer will be
$$2\lim_{x \rightarrow 0}\left(\frac{1}{x^2} - \frac{1}{x\sin x}\right)$$
$$=2\lim_{x \rightarrow 0}\left(\frac{x}{\sin x}\right)\left(\frac{\sin x - x}{x^3}\right)$$
$$=2\lim_{x \rightarrow 0}\frac{\sin x - x}{x^3}$$
An elementary calculus way of showing this limit is $-{1 \over 6}$ can be found here.
So the overall answer is $-\frac{1}{3}$.
A: sorry about to answer my Question but Zarrax's way and her comment lead me to the answer 
$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{\sin\frac{x}{3}-\frac{x}{3}}{(\frac{x}{3})^3}$$
$$L=\lim_{x\rightarrow 0}\frac{\sin(x)-x}{x^3}=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}$$
$$L=\lim_{x\rightarrow 0}\frac{3\sin(\frac{x}{3})-4\sin^3(\frac{x}{3})-x}{x^3}=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}$$
$$L=\lim_{x\rightarrow 0}\frac{3(\sin(\frac{x}{3})-\frac{x}{3})-4\sin^3(\frac{x}{3})}{27(\frac{x}{3})^3}=\frac{3L}{27}-\frac{4}{27}$$
$$L=\frac{3L}{27}-\frac{4}{27}$$
$$L=-\frac{1}{6}$$
$$2L=-\frac{1}{3}=\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$
thanks alot Zarrax
thanks for every one
A: We will use that $\lim\limits_{x\to0}\frac{\sin(x)}{x}=1$, which can be shown geometrically.
First note that
$$
\begin{align}
\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}
&=\frac{\sin^2(x)-x^2}{x^2\sin^2(x)}\\
&=\frac{\sin^2(x)-x^2}{x^4}\left(\frac{\sin(x)}{x}\right)^{-2}\\
&=\frac{\sin(x)-x}{x^3}\left(\frac{\sin(x)}{x}+1\right)\left(\frac{\sin(x)}{x}\right)^{-2}\tag{1}
\end{align}
$$
Next, we have
$$
\begin{align}
\frac{\sin(x/2^{k-1})-2\sin(x/2^k)}{x^3}
&=\frac{2\sin(x/2^k)(\cos(x/2^k)-1)}{x^3}\\
&=\frac{\sin(x/2^k)}{x^3}\frac{2(\cos^2(x/2^k)-1)}{\cos(x/2^k)+1}\\
&=-2^{-3k}\left(\frac{\sin(x/2^k)}{x/2^k}\right)^3\frac2{\cos(x/2^k)+1}\tag{2}
\end{align}
$$
Therefore,
$$
\begin{align}
\lim_{x\to0}\frac{\sin(x)-x}{x^3}
&=\lim_{x\to0}\sum_{k=1}^\infty2^{k-1}\frac{\sin(x/2^{k-1})-2\sin(x/2^k)}{x^3}\\
&=\lim_{x\to0}\sum_{k=1}^\infty-2^{-2k-1}\left(\frac{\sin(x/2^k)}{x/2^k}\right)^3\frac2{\cos(x/2^k)+1}\\
&=\sum_{k=1}^\infty-2^{-2k-1}\\
&=-\frac16\tag{3}
\end{align}
$$
Plugging $(3)$ into $(1)$, we get
$$
\begin{align}
\lim_{x\to0}\left(\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}\right)
&=-\frac16\cdot2\cdot1^2\\
&=-\frac13\tag{4}
\end{align}
$$
A: Let $I = \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)$, we have:
$$\begin{align} 
I = & \lim_{x\to 0}\left(\frac{1}{(2x)^2} - \frac{1}{\sin^2(2x)}\right)\\
  = & \lim_{x\to 0} \left(\frac{1}{4 x^2} - \frac{1}{4 \sin^2 x\cos^2 x}\right)\\
\implies  4I = & \lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)\\
\implies  3I = & \lim_{x\to 0}\left\{\left(\frac{1}{x^2} - \frac{1}{\sin^2 x\cos^2 x}\right)
                             -\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)\right\}\\
=&-\lim_{x\to 0} \frac{1-\cos^2 x}{\sin^2 x\cos^2 x} = -\lim_{x\to 0}\frac{1}{\cos^2 x} = -1\\
\implies I = & -\frac{1}{3}
\end{align}$$
