Question

Evaluate:
$\rm \displaystyle \lim_{x\to 0}\ \frac{2\sin x-\sin 2x}{x^{3}}$

Answer 1

See the image and ask if there is any problem.

Answer 2

we have to evaluate, $\rm \displaystyle \lim_{x\to 0}\ \frac{2\sin x-\sin 2x}{x^{3}}$

first of all, we have to check the form of limit. i.e., 0/0, ∞/∞, ∞ - ∞, 0.∞, 1^∞, 0^∞, ∞^0.

putting, x = 0 in function, f(x) = {2sinx - sin2x}/x³

so, f(0) = {2sin(0) - sin2(0)}/(0)³ = 0/0, it is the form of limit.

now, $\rm \displaystyle \lim_{x\to 0}\ \frac{2\sin x-\sin 2x}{x^{3}}$

= $\displaystyle \lim_{x\to 0}\ \frac{2sinx-2sin x cos x}{x^{3}}$

= $\displaystyle \lim_{x\to 0}\ \frac{2sin x(1-cosx)}{x^{3}}$

= 2$\displaystyle \lim_{x\to 0}\ \frac{sin x\left(2sin^{2}\frac{x}{2}\right)}{x^{3}}$

=4$\displaystyle \lim_{x\to 0}\ \frac{sinx}{x}\frac{sin^{2}\frac{x}{2}}{\left(\frac{x}{2}\right)^2\times4}$

= 4 × 1 × 1/4

= 1