# Question cd30e

Sep 1, 2015

${\lim}_{x \to 0} \frac{\arcsin x}{4 x} = \frac{1}{4}$

#### Explanation:

First, you need to know that

${\lim}_{x \to 0} \arcsin x = 0 \text{ }$

This means that if you try to evaluate this limit by replacing $x$ with zero, you will get

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{\lim}_{x \to 0} \frac{\arcsin 0}{4 \cdot 0} = \frac{0}{0}}}}$

As you know, $\frac{0}{0}$ is an indeterminate form, so you need to use L'Hopital's Rule to find the limit.

L'Hopital's Rule tells you that if the limit

${\lim}_{x \to a} \frac{{f}^{'} \left(x\right)}{{g}^{'} \left(x\right)}$

exists, and provided that ${g}^{'} \left(x\right) \ne 0$, then you have

$\textcolor{b l u e}{{\lim}_{x \to 0} \frac{f \left(x\right)}{g \left(x\right)} = {\lim}_{x \to a} \frac{{f}^{'} \left(x\right)}{{g}^{'} \left(x\right)}}$

In your case, you have $a = 0$. Provided that you know that

$\frac{d}{\mathrm{dx}} \left(\arcsin x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

then you have

${\lim}_{x \to 0} \frac{\arcsin x}{4 x} = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left(\arcsin x\right)}{\frac{d}{\mathrm{dx}} \left(4 x\right)} = \frac{\frac{1}{\sqrt{1 - {x}^{2}}}}{4} = \frac{1}{4} \cdot \frac{1}{\sqrt{1 - {x}^{2}}}$

Now you can evaluate this limit for $x \to 0$ by replacing $x$ with zero to get

${\lim}_{x \to 0} \frac{\arcsin x}{4 x} = \frac{1}{4} \cdot \frac{1}{\sqrt{1 - {0}^{2}}} = \frac{1}{4} \cdot 1 = \textcolor{g r e e n}{\frac{1}{4}}$

Sep 1, 2015

${\lim}_{x \rightarrow 0} \frac{\arcsin x}{4 x} = \frac{1}{4}$

#### Explanation:

This question was (originally) posted under the topic "Determining Limits Graphically". If that is the way you are doing it, look at the graph:

graph{arcsinx/(4x) [-1.411, 1.29, -0.281, 1.069]}

As $x \rightarrow 0$, it appears that $y \rightarrow \frac{1}{4}$.

Without graphing technology, we can do a substitution and use the fundamental trigonometric limit.

Let $\theta = \arcsin x$.

The $x = \sin \theta$ and as $x \rightarrow 0$, we have $\theta \rightarrow 0$

So we can rewrite the limit as:

${\lim}_{\theta \rightarrow 0} \frac{\theta}{4 \sin \theta} = \frac{1}{4} {\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin \theta} = \frac{1}{4} \left(1\right) = \frac{1}{4}$