TITLE: Simplify trig function and calculate limit $\lim\limits_{x \to 0} \frac{\tan x-\sin x}{\sin^2 x}$
QUESTION [5 upvotes]: Using the fact that $\lim\limits_{x \to 0} \frac{\sin(x)}{x}=1$, please help me show that

$$\lim\limits_{x \to 0} \frac{\tan(x)-\sin(x)}{\sin^2(x)}=0.$$

Because I am not familar with L'Hôpital's rule and Taylor's Theorem, please avoid the use of either in your solution.

REPLY [8 votes]: You could also finish Jared's first expression by doing this
$$
\begin{align}
&\lim_{x\rightarrow0}\frac{1-\cos x}{\cos x\sin x}\cdot\frac{1+\cos x}{1+\cos x}\\
=&\lim_{x\rightarrow0}\frac{1-\cos^2 x}{\cos x\sin x(1+\cos x)}\\
=&\lim_{x\rightarrow0}\frac{\sin^2 x}{\cos x\sin x(1+\cos x)}\\
=&\lim_{x\rightarrow0}\frac{\sin x}{\cos x(1+\cos x)}
\end{align}
$$
The limit of the denominator is 2 and the limit of the numerator is 0.

REPLY [5 votes]: $$\lim\limits_{x \to 0} \frac{\tan(x)-\sin(x)}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\frac{\sin(x)}{\cos (x)}-\sin(x)}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\frac{1}{\cos (x)}-1}{\sin(x)}=\lim\limits_{x \to 0} \frac{2\sin^2(\frac x2)}{2\sin(2x)}=.\lim\limits_{x \to 0} \frac x8=0$$