# 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:

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

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.

Hint:Using $$\cos(2x) = \cos^2(x)-\sin^2(x) =1-\sin^2(x)-\sin^2(x)=1-2\sin^2(x),$$ we get $$\frac{5-5\cos(2x)}{x} = 10\sin(x)\cdot\frac{\sin(x)}{x}.$$

• And I thought replacing $cos(2x)=cos^2(x)−sin^2(x)$ was useless. How wrong was I. I need to practice more, thank you very much! Jul 7, 2018 at 23:19

Remember the double-angle formula $\cos(2x)=1-2\sin^2 x$?

• You're quite welcome! Jul 7, 2018 at 23:34

Pythagoras' identity will do: $$\frac{5(1-\cos 2x)}x=\frac{5(1-\cos^2 2x)}{x(1+\cos 2x)}=\frac{5\sin^2 2x}{x(1+\cos 2x)}=5\underbrace{\frac{\sin 2x}{x}}_{\substack{\downarrow}\\2}\,\frac{\sin 2x\rlap{\:\to 0}}{(1+\cos 2x)\rlap{\:\to 2}}$$

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.