Simplify trig function and calculate limit $\lim\limits_{x \to 0} \frac{\tan x-\sin x}{\sin^2 x}$ 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.
 A: 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.
A: $$\lim\limits_{x \to 0} \frac{\tan(x)-\sin(x)}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\frac{\sin(x)}{\cos (x)}-\sin(x)}{\sin^2(x)}=\lim\limits_{x \to 0} \frac{\frac{1}{\cos (x)}-1}{\sin(x)}=\lim\limits_{x \to 0} \frac{2\sin^2(\frac x2)}{2\sin(2x)}=.\lim\limits_{x \to 0} \frac x8=0$$
A: If you multiply top and bottom by $\cot x$, you get:
$$\lim_{x\to 0}\frac{1}{\cos x}\frac{1-\cos x}{\sin x}$$
Using the taylor series for sine and cosine, we have:
$$\lim_{x\to 0}\frac{1}{1-\frac{x^2}{2!}+\ldots}\frac{\frac{x^2}{2!}+\ldots}{x-\frac{x^3}{3!}+\ldots}=\lim_{x\to 0}\frac{1}{1-\frac{x^2}{2!}+\ldots}\frac{\frac{x}{2!}+\ldots}{1-\frac{x^2}{3!}+\ldots}=0$$
after canceling an $x$ and plugging in $0$.
A: The answer is zero, because the numerator is an odd function whose linear terms cancel, so it behaves as $a x^3$ as $x \to 0$, for some $a$, while the denominator behaves as $x^2$ in that limit.  The latter point should be clear from your statement that $\sin{x}/x$ approaches $1$ as $x \to 0$.  The former point may be reasoned as follows:
$$\sin{x} \sim x + \text{higher-order terms}$$
$$\tan{x} \sim x+ \text{higher-order terms}$$
$\tan{x}-\sin{x}$ is odd.  Therefore there are no even terms in an expansion.  Therefore, the lowest-order term is $a x^3$ for some $a$.
ADDENDUM
You can also use simple trig identities, e.g.,
$$\frac{\tan{x}-\sin{x}}{\sin^2{x}} = \frac{\tan{x}}{1+\cos{x}}$$
and clearly, the limiting value of the RHS as $x \to 0$ is zero.
A: Do you know the derivatives of trig functions? If so: first, write $\tan x = \sin x/\cos x$. Then simplify the resulting fraction and multiply by $1 = \frac{x - 0}{x - 0}$, and use the limit you already know and a particular interpretation of what remains to see that the limit is indeed $0$. To start you off:
\begin{align*}
\lim_{x\to 0}{\frac{\tan(x)-\sin(x)}{\sin^2x}} &= \lim_{x\to 0}{\frac{\sin(x)/\cos(x)-\sin(x)}{\sin^2x}}\\
&= \lim_{x\to 0}{\frac{\sin(x)-\cos(x)\sin(x)}{\cos(x)\sin^2x}}\\
&= \lim_{x\to 0}\frac{1}{\cos x}\cdot\lim_{x\to 0}{\frac{1-\cos(x)}{\sin(x)}}\\
&= \lim_{x\to 0}{\frac{1-\cos(x)}{\sin(x)}}.
\end{align*}
Can you take it from here?
A: Hint: If $\cos(x)\ne1$,
$$
\begin{align}
\frac{\tan(x)-\sin(x)}{\sin^2(x)}
&=\frac{\tan(x)(1-\cos(x))}{1-\cos^2(x)}\\
&=\frac{\tan(x)}{1+\cos(x)}\\
&=\frac{\sin(x)}{\cos(x)+\cos^2(x)}\\
&\to\frac02
\end{align}
$$
