Solving Limit Problem without L'Hospital Rule $\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}$ How can we find the limit $\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}$
I keep getting DNE as my answer. The answer is supposed to be 0.
My Work:
$\lim_{x\to 0} \frac{\cos(x)-1}{1}*\frac{1}{\sin(2x)}$
$\lim_{x\to 0} \frac{\cos(x)-1}{1}*\frac{1}{\sin(2x)}*\frac{2x}{2x}$
$\lim_{x\to 0} \frac{\cos(x)-1}{1}*\frac{1}{2x}$
$\lim_{x\to 0} \frac{\cos(x)-1}{2x}$
$\lim_{x\to 0} \frac{\cos(0)-1}{2(0)}$
$\frac{0}{0}$
I am unsure what I am doing wrong. This is a problem in my HW. When I solve using L'Hospital Rule I get 0 and the answer key is in agreement with it. However we are explicitly told to solve without use of L'Hospital Rule.
Thank you
 A: HINT
Multiply and divide by $\cos(x) + 1$:
\begin{align*}
\lim_{x\to 0}\frac{\cos(x) - 1}{\sin(2x)} & = \lim_{x\to 0}\frac{\cos^{2}(x) - 1}{2\sin(x)\cos(x)}\times\frac{1}{\cos(x) + 1}\\\\
& = \lim_{x\to 0}-\frac{\sin^{2}(x)}{2\sin(x)\cos(x)}\times\frac{1}{\cos(x) + 1}\\\\
& = \lim_{x\to 0}-\frac{\sin(x)}{2\cos(x)}\times\frac{1}{\cos(x) + 1}
\end{align*}
Can you take it from here?
A: $$
\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}
$$
Also you can use the Taylor series:
$$
\cos(x)=\left(1-\frac{x^2}{2}+\frac{x^4}{24}-....\right),\tag{1}
$$
$$
\sin(2x)=\left(2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}+\frac{(2x)^7}{7!}+....\right),\tag{2}
$$
Therefore,
$$
\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}=\lim_{x\to{0}}\frac{\left(1-\frac{x^2}{2}+\frac{x^4}{24}-....\right)-1}{\left(2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}+\frac{(2x)^7}{7!}+....\right)}=
$$
$$
=\lim_{x\to{0}}\frac{x\left(-\frac{x}{2}+\frac{x^3}{24}-....\right)}{x\left(2-\frac{(2x)^2}{3!}+\frac{(2x)^4}{5!}+\frac{(2x)^6}{7!}+....\right)}=\lim_{x\to{0}}\frac{0}{2}=0.
$$
Therefore,
$$
\lim_{x\to 0} \frac{\cos(x)-1}{\sin(2x)}=0.
$$
