Evaluate $\lim_{x \to 0} \frac{2 - \cos(3x) - \cos(4x)}{x}$ How do you evaluate $$\lim_{x \to 0} \frac{2 - \cos(3x) - \cos(4x)}{x}?$$ I have looked at this problem for a while and cannot think of a way of algebraically determining this limit although I know one must exist.
 A: An alternative way to find the limit is the following:
$\lim_{x\to0}\frac{2-\cos(3x)-\cos(4x)}{x}=\lim_{x\to0}\bigg(\frac{1-\cos(3x)}{x}+\frac{1-\cos(4x)}{x}\bigg)$
$=\lim_{x\to0}\bigg(3\frac{1-\cos(3x)}{3x}+4\frac{1-\cos(4x)}{4x}\bigg)$
$=\lim_{x\to0}\bigg(3\frac{1-\cos^{2}(3x)}{3x}\cdot\frac{1}{1+\cos(3x)}+4\frac{1-\cos^{2}(4x)}{4x}\cdot\frac{1}{1+\cos(4x)}\bigg)$
$=\lim_{x\to0}\bigg(3\frac{\sin(3x)}{3x}\cdot\frac{\sin(3x)}{1+\cos(3x)}+4\frac{\sin(4x)}{4x}\cdot\frac{\sin(4x)}{1+\cos(4x)}\bigg)$
A: Define $f\colon \mathbb R \to\mathbb R, x\mapsto \cos(3x)+\cos(4x)$ and note that  $$\begin{align} 
\lim \limits_{x\to 0}\left(\dfrac{\cos(3x)+\cos(4x)-2}{x}\right)&=\lim \limits_{x\to 0}\left(\dfrac{\cos(3x)+\cos(4x)-(\cos(3\cdot 0)+\cos(4\cdot 0))}{x-0}\right)\\
&=\lim \limits_{x\to 0}\left(\dfrac{f(x)-f(0)}{x-0}\right)\\
&=f'(0).
\end{align}$$
A: There is a simple solution if you remember that, around $y=0$ the Taylor series of $\cos(y)$ is simply $(1-\frac {y^2}{2})$. Then $$2 -\cos(3x)-\cos(4x) \simeq 2 - (1-\frac {9x^2}{2})- (1-\frac {16x^2}{2})=\frac {25x^2}{2}$$ I am sure that you can take from here.
