# Limit without L'Hopital's rule: $\lim_{x\to0} \frac{1-\cos^3 x}{x\sin2x}$

How can I solve the following problem without the use of the L'Hopitals's rule?

$$\lim_{x\to0} \frac{1-\cos^3(x)}{x\sin{(2x)}}$$

Hint:

We have that $1-\cos^3x = (1-\cos x)(1+\cos x + \cos^2 x)$ and $x\sin 2x = 2x\sin x \cos x$, so by limit arithmetic $$\lim_{x \to 0} \frac{1-\cos^3x}{x\sin 2x} = \lim_{x\to0} \frac{(1-\cos x)(1+\cos x + \cos^2 x)}{2x\sin x \cos x} = \frac{3}{2}\lim_{x \to 0}\frac{1-\cos x}{x\sin x}$$ From here, see if you can get a numerator in terms of $\sin x$.

Hint: Factor $1-\cos^3 x$ and note that $1-\cos x=2\sin^2(x/2)$.

we have $-\frac{(\cos(x)-1)(\cos(x)^2+\cos(x)+1)}{2x\sin(x)\cos(x)}=-\frac{(\cos(x)^2-1)(\cos(x)^2+\cos(x)+1)}{2x\sin(x)\cos(x)(1+\cos(x))}=\frac{(\sin(x)(\cos(x)^2+\cos(x)+1)}{2x\cos(x)(1+\cos(x))}$ and the searched limit is $\frac{3}{4}$

$$\frac{1-\cos^3(x)}{x\sin(2x)}=\frac{(1-\cos(x))(1+\cos(x)+\cos^2(x))}{2x\sin(x)\cos(x)}$$ $$=\frac{2\sin^2(x/2)(1+\cos(x)+\cos^2(x))}{4x\sin(x/2)\cos(x/2)\cos(x)}$$ $$=\frac{1}{4}\frac{\sin(x/2)}{x/2}\frac{1+\cos(x)+\cos^2(x)}{\cos(x/2)\cos(x)}$$

The limit is 3/4.

$$\frac{1-\cos^3(x)}{x\sin 2x}= \frac{1-(1-3x^2/2+O(x^4))}{2x^2-O(x^4)}=\frac{3x^2/2+O(x^4)}{2x^2+O(x^4)}=\frac 34$$

Yes, you can: remember the two basic limits $$\lim_{x\to0}\frac{\sin x}{x}=1, \qquad \lim_{x\to0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$ and rewrite your limit as $$\lim_{x\to0}\frac{1}{2}\frac{1-\cos x}{x^2}\frac{2x}{\sin2x}(1+\cos x+\cos^2x)$$

• The second one is not $1/2$. It is $0$. – Jack Oct 7 '16 at 3:07
• @Jack Right. I forgot a two! – egreg Oct 7 '16 at 8:04