Compute the limit:

Question

$$ \lim_{x \geq 0}{\frac{1-\cos^3(x)}{x\sin(2x)}}$$ I tried: $$\lim_{x \geq 0}{{\frac{1-\cos^2(x)*\cos(x)}{2*x*\sin^2(x)\cos^2(x)}}}$$ $$\lim_{x \geq 0}{\frac{\cos^2x+\sin^2x-\cos^2(x)\cos(x)}{2*x*\sin^2x*\cos^2(x)}}$$ $$\lim_{x \geq 0}{\frac{\cos^2(x)+\sin^2(x)-(1-\sin^2(x))\cos(x)}{2*x*\sin^2(x)*\cos^2(x)}}$$ $$\lim_{x \geq 0}{\frac{\cos^2(x)+\sin^2(x)-\cos(x)+\sin^2(x)\cos(x)}{2*x*\sin^2(x)*\cos^2(x)}}$$ Somewhere I over-complicated and I don't know how to continue or the other route that I should take...

Answer 1

We have that

$${\frac{1-\cos^3(x)}{x\sin(2x)}}={\frac{1-\cos(x)}{x^2}}\cdot(1+\cos x+\cos^2(x))\cdot\frac12{\frac{2x}{\sin(2x)}}$$

then refer to standard limits

• $\frac{1-\cos(x)}{x^2}\to \frac12$
• $\frac{2x}{\sin(2x)}\to 1$

and simply

• $1+\cos x+\cos^2(x)\to 3$

Answer 2

Hint: Write $$\frac{(1-\cos(x))(1+\cos(x))(\cos^2(x)+\cos(x)+1)}{2x\sin(x)\cos(x)(1+\cos(x))}$$

Answer 3

\begin{align}\lim_{x \to 0}\frac{1-\cos^3x}{x \sin (2x)}&= \lim_{x \to 0}\frac{1-\left( 1-\frac{x^2}2\right)^3}{2x^2}\\ &= \lim_{x \to 0} \frac{1-(1-\frac{3x^2}2)}{2x^2} \\ &= \lim_{x \to 0} \frac{\frac32x^2}{2x^2}\\ &= \frac34\end{align}

Answer 4

$$\lim_{x\rightarrow0}\dfrac{1-\cos^3x}{x\sin(2x)}$$ Apply L'Hopital's Rule$$\lim_{x\rightarrow0}\dfrac{1-\cos^3x}{x\sin(2x)}=\lim_{x\rightarrow0}\dfrac{3\cos^2x\sin x}{\sin(2x)+2x\cos(2x)}$$ Again apply L'Hopital's Rule $$\lim_{x\rightarrow0}\dfrac{3(-\sin(2x)\sin x+\cos^3x)}{4\cos(2x)-4x\sin(2x)}=\dfrac34$$

Answer 5

Let's write the expression under limit as $$\frac{1-\cos ^3x}{1-\cos x} \cdot\frac{1-\cos x} {x^2}\cdot\frac{2x}{\sin 2x}\cdot\frac{1}{2}$$ The first fraction tends to $3$, the second one tends to $1/2$, and the third one tends to $1$ and therefore the desired limit is $3/4$.