Find limit of the expression $$\lim_{x \to 0}\frac{\cos(x) - 8x\sin(x/2) - \cos(3x)}{x^4}$$
I think I should replace by equivalent, such as $\sin(x)$ ~ $x$, but got nothing.
Thank you for answers, but what about solving without using L'Hôpital's rule?
 A: $\cos(x)-\cos(3x) = 2\sin(x)\sin(2x) = 4\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\sin(2x)$, hence we want to compute:
$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)-2x}{x^3}=2\cdot\lim_{x\to 0}\frac{\sin(2x)-2x}{x^3}=16\cdot\lim_{z\to 0}\frac{\sin(z)-z}{z^3} $$
that equals $-\frac{8}{3}$, by applying twice De l'Hopital theorem, then subtract:
$$ \lim_{x\to 0}\frac{4\sin\left(\frac{x}{2}\right)}{x}\cdot\frac{\sin(2x)(1-\cos\frac{x}{2})}{x^3}=4\cdot\lim_{x\to 0}\frac{\sin(2x)\sin^2\left(\frac{x}{4}\right)}{x^3}=\frac{1}{2}. $$
The given limit is so $-\frac{8}{3}-\frac{1}{2}=\color{red}{\Large -\frac{19}{6}}$.

To prove the crucial part, i.e. $\lim_{x\to 0}\frac{x-\sin(x)}{x^3}=\frac{1}{6}$, without derivatives, you may assume that the limit just exists and equals $L$. Then:
$$ L = \lim_{x\to 0}\frac{2x-\sin(2x)}{8x^3} = \lim_{x\to 0}\frac{x-\sin(x)\cos(x)}{4x^3}=\frac{L}{4}+\lim_{x\to 0}\frac{\sin(x)\sin^2\left(\frac{x}{2}\right)}{2x^3} $$
and that leads to $L=\frac{L}{4}+\frac{1}{8}$, from which $L=\frac{1}{6}$.
A: You need to consider all three terms of the numerator at the same time, as
they are all needed to cancel.
You know that the numerator has a power series expansion, so let
$n$ be the numerator. Then, either by direct computation, or by expanding each
term, we have
$n(0) = n'(0) = n''(0) = n'''(0) = 0$, $n''''(0) = - {19 \over 6}$.
Hence $n(x) = - {19 \over 6}x^4 + x^5 r(x)$, where $r$ is bounded near zero, hence we see the limit is $- {19 \over 6}$.
