# Question #cd30e

##### 2 Answers

#### Explanation:

First, you need to know that

#lim_(x -> 0)arcsinx = 0" "#

This means that if you try to evaluate this limit by replacing

#color(red)(cancel(color(black)(lim_(x->0)(arcsin0)/(4 * 0) = 0/0)))#

As you know, **L'Hopital's Rule** to find the limit.

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

#lim_(x->a)(f^'(x))/(g^'(x))#

exists, and provided that

#color(blue)(lim_(x->0)(f(x))/(g(x)) = lim_(x->a)(f^'(x))/(g^'(x)))#

In your case, you have

#d/dx(arcsinx) = 1/sqrt(1-x^2)#

then you have

#lim_(x->0)(arcsinx)/(4x) = lim_(x->0)(d/dx(arcsinx))/(d/dx(4x)) = (1/sqrt(1-x^2))/4 = 1/4 * 1/sqrt(1-x^2)#

Now you can evaluate this limit for

#lim_(x->0)(arcsinx)/(4x) = 1/4 * 1/sqrt(1 - 0^2) = 1/4 * 1 = color(green)(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

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

Let

The

So we can rewrite the limit as: