How to evaluate $\lim _{x\to 0}\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}$ without using L'Hospital's rule? I need to evaluate the following limit without using L'Hopital's rule:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}\right)$$
I thought the best way was to separate it in two limits:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)}{x}\right)+\lim_{x\to \:0}\left(\frac{\sin\left(4x\right)}{x}\right)$$
Considering that $\lim\,_{x\to \:0}\left(\frac{\sin x}{x}\right)=1$ we can easily know that the second limit is $4$. I also know the result of the original limit is $4$, so the result of the first limit in the second line needs to be $0$ $(0+4=4)$.
The issue is that I can not figure out how to remove the $x$ from the denominator so I can avoid the indeterminate form. I already tried replacing $\cos(2x)$ with $\cos^2 x-\sin^2 x$, but it seems to be useless.
So, to summarize everything, my problem is how to evaluate this limit without using L'Hospital's rule:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)}{x}\right).$$
 A: Hint: Use these two facts and a little bit of algebra:
$$\lim_{x \to 0} \frac{1-\cos(x)}{x} = 0$$
$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
Note that $$\lim_{x \to 0} 10 \times \frac{1-\cos(2x)}{2x} = 0$$
The most straightforward and rigorous way to see why the first relation holds is to use the Taylor series for cosine. A non-rigorous way of seeing the first relation holds, if we accept the second relation, besides using the double-angle formula and similar trigonometric relations as pointed out by others, is this non-rigorous but more or less intuitive argument:
When $x$ is small, we know that:
$$\sin(x) \approx x $$
$$\cos(x)=\int -\sin(x) \approx -x^2/2+C$$
If $x=0$, then $\cos(x)=1$, hence $C=1$.
This shows that $1-\cos(x)=x^2/2$ when $x$ is small and this proves the first relation.
A: Hint:Using $$\cos(2x) = \cos^2(x)-\sin^2(x) =1-\sin^2(x)-\sin^2(x)=1-2\sin^2(x),$$
we get $$\frac{5-5\cos(2x)}{x} = 10\sin(x)\cdot\frac{\sin(x)}{x}.$$
A: Remember the double-angle formula $\cos(2x)=1-2\sin^2 x$?
A: Pythagoras' identity will do:
$$\frac{5(1-\cos 2x)}x=\frac{5(1-\cos^2 2x)}{x(1+\cos 2x)}=\frac{5\sin^2 2x}{x(1+\cos 2x)}=5\underbrace{\frac{\sin 2x}{x}}_{\substack{\downarrow}\\2}\,\frac{\sin 2x\rlap{\:\to 0}}{(1+\cos 2x)\rlap{\:\to 2}}$$
A: Just another way using Taylor series.
$$\cos(x)=1-\frac{x^2}{2}+O\left(x^4\right)\implies \cos(2x)=1-2 x^2+O\left(x^4\right)$$
$$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)\implies \sin(4x)=4 x-\frac{32 x^3}{3}+O\left(x^4\right)$$
$$5-5\cos\left(2x\right)+\sin\left(4x\right)=4 x+10 x^2-\frac{32 x^3}{3}+O\left(x^4\right)$$
$$\frac{5-5\cos\left(2x\right)+\sin\left(4x\right)}{x}=4+10 x-\frac{32 x^2}{3}+O\left(x^3\right)$$ which shows the limit and how it is approached.
