<s>[INST] 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:
$$\lim _{x\to 0}\left(\frac{5-5\cos\left(2x\right)}{x}\right).$$
 [/INST] 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.
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