Limits of trig functions How can I find the following problems using elementary trigonometry?
$$\lim_{x\to 0}\frac{1−\cos x}{x^2}.$$
$$\lim_{x\to0}\frac{\tan x−\sin x}{x^3}.
$$
Have attempted trig identities, didn't help. 
 A: HINT :
\begin{align}
\lim_{x\to 0}\frac{1−\cos x}{x^2}&=\lim_{x\to 0}\frac{1−\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\\
&=\lim_{x\to 0}\frac{1−\cos^2 x}{x^2(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin x\cdot\sin x}{x\cdot x\cdot(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac{1}{1+\cos x}\\
&=1\cdot1\cdot\frac1{1+1}
\end{align}
and
\begin{align}
\lim_{x\to 0}\frac{\tan x−\sin x}{x^3}&=\lim_{x\to 0}\frac{\tan x−\sin x}{x^3}\cdot\frac{\cos x}{\cos x}\\
&=\lim_{x\to 0}\frac{\sin x(1−\cos x)}{x^3\cos x}\\
&=\lim_{x\to 0}\frac{\sin x(1−\cos x)}{x^3\cos x}\cdot\frac{1+\cos x}{1+\cos x}\\
&=\lim_{x\to 0}\frac{\sin^3 x}{x^3\cos x(1+\cos x)},\\
\end{align}
where $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1.$
A: Using $\cos x = 1-2\sin^2\frac x2$ gives you $$\frac{1-\cos x}{x^2} = \frac{2\sin^2\frac x2}{x^2} = \frac12\left(\frac{\sin\frac x2}{\frac x2}\right)^2.$$
Using similar identities will help with the second limit as well.
A: L'hopital's rule is useful in this case, since the limits are of indeterminate form $0/0$. Then, taking the derivative of the numerator and denominator:
$$
\lim_{x\to 0}\frac{1−\cos x}{x^2} = \lim_{x\to 0}\frac{\sin x}{2x} = \frac{1}{2}\lim_{x\to 0} \frac{\sin x}{x}.
$$
which should look more familiar. The second limit is the same deal, except you apply L'hopital's rule 3 times (until the denominator is a constant). See if you can work it out yourself!
