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:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)}{x}\right).$$
Best Answer
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.