Question

lim x tends to 0 (1-cos cube x)/xsin2x

Answer 1

Limit x tends to zero sinx by x = 1

Answer 2

Answer:

3/4

Step-by-step explanation:

$\displaystyle\frac{1-\cos^3x}{x\sin2x}=\frac{(1-\cos x)(1+\cos x+\cos^2x)}{x\sin2x}\\\\{}\qquad=\frac{(1-\cos^2x)(1+\cos x+\cos^2x)}{x\sin2x(1+\cos x)}\\\\{}\qquad=\frac{\sin^2x(1+\cos x+\cos^2x)}{x\sin2x(1+\cos x)}\\\\{}\qquad=\frac{\sin x}{x}\cdot\frac{\sin x}{x}\cdot\frac{2x}{\sin2x}\cdot\frac{1+\cos x+\cos^2x}{2(1+\cos x)}$

So...

$\displaystyle\lim_{x\rightarrow0}\frac{1-\cos^3x}{x\sin2x}\\\\=\left(\lim_{x\rightarrow0}\frac{\sin x}{x}\right)\left(\lim_{x\rightarrow0}\frac{\sin x}{x}\right)\left(\lim_{x\rightarrow0}\frac{2x}{\sin2x}\right)\left(\lim_{x\rightarrow0}\frac{1+\cos x+\cos^2x}{2(1+\cos x)}\right)\\\\=1\times1\times1\times\frac34\\\\=\frac34$