Question

Evaluate limit x tends to zero 10 x minus sin x by x cube

Answer 1

Step-by-step explanation:

lt. 10x-sinx/x cube

x--)0

0/0square-1

-1

Answer 2

The value of $\lim_{x \to 0} \dfrac{\tan x-\sin x}{x^3}$ is equal to $\dfrac{1}{2}$.

Step-by-step explanation:

The complete question:

$\lim_{x \to 0} \dfrac{\tan x-\sin x}{x^3}$

To find, the value of $\lim_{x \to 0} \dfrac{\tan x-\sin x}{x^3}$ = ?

∴ $\lim_{x \to 0} \dfrac{\tan x-\sin x}{x^3}$

$=\lim_{x \to 0} \dfrac{\dfrac{\sin x}{\cos x} -\sin x}{x^3}$

$=\lim_{x \to 0} \dfrac{\sin x(1-\cos x)}{\cos x.x^3}$

$=\lim_{x \to 0} \dfrac{\dfrac{\sin x}{\cos x} 2\sin^2 \dfrac{x}{2}}{x^3}$

$=\lim_{x \to 0} 2\dfrac{\tan x}{x}(\dfrac{1}{4} )(\dfrac{\sin \dfrac{x}{2}}{\dfrac{x}{2}})^2$

We know that,

$\lim_{x \to 0} \dfrac{\tan x}{x}$ = 1 and

$\lim_{x \to 0} \dfrac{\sin x}{x}$ = 1

= $2\times \dfrac{1}{4}$

= $\dfrac{1}{2}$

∴ The value of $\lim_{x \to 0} \dfrac{\tan x-\sin x}{x^3}$ = $\dfrac{1}{2}$

Thus, the value of $\lim_{x \to 0} \dfrac{\tan x-\sin x}{x^3}$ is equal to $\dfrac{1}{2}$.