# Lim x->0 (tan x- sin x)/x^3=?

## a. 0 b. 1 c. 1/2 d. -1 e. -1/2

Nov 9, 2017

${\lim}_{x \to 0} \frac{\tan x - \sin x}{x} ^ 3 = \frac{1}{2}$

#### Explanation:

Transform the function in this way:

$\frac{\tan x - \sin x}{x} ^ 3 = \frac{1}{x} ^ 3 \left(\sin \frac{x}{\cos} x - \sin x\right)$

$\frac{\tan x - \sin x}{x} ^ 3 = \frac{1}{x} ^ 3 \left(\frac{\sin x - \sin x \cos x}{\cos} x\right)$

$\frac{\tan x - \sin x}{x} ^ 3 = \sin \frac{x}{x} ^ 3 \frac{1 - \cos x}{\cos} x$

$\frac{\tan x - \sin x}{x} ^ 3 = \left(\sin \frac{x}{x}\right) \left(\frac{1 - \cos x}{x} ^ 2\right) \left(\frac{1}{\cos} x\right)$

We can use now the well known trigonometric limit:

${\lim}_{x \to 0} \sin \frac{x}{x} = 1$

and using the trigonometric identity:

${\sin}^{2} \alpha = \frac{1 - \cos 2 \alpha}{2}$

we have:

${\lim}_{x \to 0} \frac{1 - \cos x}{x} ^ 2 = {\lim}_{x \to 0} \frac{2 {\sin}^{2} \left(\frac{x}{2}\right)}{x} ^ 2 = \frac{1}{2} {\lim}_{x \to 0} {\left(\sin \frac{\frac{x}{2}}{\frac{x}{2}}\right)}^{2} = \frac{1}{2}$

While the third function is continuous so:

${\lim}_{x \to 0} \frac{1}{\cos} x = \frac{1}{1} = 1$

and we can conclude that:

${\lim}_{x \to 0} \frac{\tan x - \sin x}{x} ^ 3 = {\lim}_{x \to 0} \left(\sin \frac{x}{x}\right) \left(\frac{1 - \cos x}{x} ^ 2\right) \left(\frac{1}{\cos} x\right) = 1 \times \frac{1}{2} \times 1 = \frac{1}{2}$

graph{(tanx-sinx)/x^3 [-1.25, 1.25, -0.025, 1]}