TITLE: Find the limit without use of L'Hôpital or Taylor series: $\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2 x}\right)$
QUESTION [9 upvotes]: Find the limit without the use of L'Hôpital's rule or Taylor series
$$\lim \limits_{x\rightarrow 0} \left(\frac{1}{x^2}-\frac{1}{\sin^2x}\right)$$

REPLY [6 votes]: 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
$$
\begin{align}
\lim_{x\to0}\left(\frac1{\vphantom{()^2}x^2}-\frac1{\sin^2(x)}\right)
&=-\frac16\cdot2\cdot1^2\\
&=-\frac13\tag{4}
\end{align}
$$