# Evaluate: lim_(x→0) (arcsinx-x)/x^3?

Feb 7, 2018

$0.$

#### Explanation:

We know that, $\left(1\right) : {\lim}_{\theta \to 0} \frac{\theta}{\sin} \theta = 1$.

$\left(2\right) : \sin 3 \theta = 3 \sin \theta - 4 {\sin}^{3} \theta$.

Let, $a r c \sin x = \theta , \text{ so that, } x = \sin \theta$.

Also, as $x \to 0 , \theta \to 0 , \mathmr{and} \sin \theta \to 0 , \text{ as well}$.

$\therefore \text{The Reqd. Lim.} = {\lim}_{x \to 0} \frac{a r c \sin x - x}{x} ^ 3$,

$= {\lim}_{\theta \to 0} \frac{\theta - \sin \theta}{\sin} ^ 3 \theta$,

$= \lim \frac{\theta - \sin \theta}{\frac{1}{4} \left(3 \sin \theta - \sin 3 \theta\right)} \ldots \ldots \left[\because , \left(2\right)\right]$,

$= \lim \frac{4 \left(\theta - \sin \theta\right)}{3 \sin \theta - \sin 3 \theta}$,

$= \lim \frac{4 \sin \theta \left(\frac{\theta}{\sin} \theta - 1\right)}{\sin \theta \left(3 - {\sin}^{2} \theta\right)}$,

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

$= \frac{1 - 1}{3 - {0}^{2}} \ldots \ldots \ldots \ldots \left[\because , \left(1\right)\right]$.

$\Rightarrow \text{ The Reqd. Lim} = 0$.

Feb 7, 2018

$\frac{1}{6}$

#### Explanation:

Substituting $0$ in the given function we get a $\frac{0}{0}$ indeterminate form.
We can use the L'Hospital's Rule,
lim_(x→0) (1/sqrt(1-x^2)-1)/(3x^2)

lim_(x→0) ((1-sqrt(1-x^2))/sqrt(1-x^2))/(3x^2)

lim_(x→0) (1-sqrt(1-x^2))/(3x^2) . lim_(x -> 0) (1/sqrt(1-x^2))

Evaluating second limit which is $1$ and applying L'Hospital's Rule to first limit,

lim_(x→0) (-(-2x)/(2sqrt(1-x^2)))/(6x)

lim_(x→0) 1/(6(sqrt(1-x^2))

Substituting $0$ and we get $\frac{1}{6}$