Question

Question #cd30e

Answer 1

`lim_(x -> 0)(arcsinx)/(4x) = 1/4`

Explanation:

First, you need to know that

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

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

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

As you know, `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->a)(f^'(x))/(g^'(x))`

exists, and provided that `g^'(x)!=0`, then you have

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

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

`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 `x->0` by replacing `x` with zero to get

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

Answer 2

`lim_(xrarr0) (arcsinx)/(4x) = 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 `xrarr0`, it appears that `yrarr1/4`.

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

Let `theta = arcsinx`.

The `x = sin theta` and as `xrarr0`, we have `theta rarr 0`

So we can rewrite the limit as:

`lim_(thetararr0) theta/(4sintheta) = 1/4lim_(thetararr0) theta/(sintheta) = 1/4 (1) = 1/4`#