# Homework Help: Lim of trig functions. Does it exist?

1. May 30, 2013

### mathgeek69

1. Does the limit exist of the following:

lim as x→ 1- ((cos^-1(x))/(1-x))

2. Relevant equations:

3. The attempt at a solution:

lim as x→ 1- ((cos^-1(x))/(1-x))
= lim as x→ 1- (cos^-1(x))/ lim as x→ 1-(1-x)

Let y = 1-x

lim as y→0 (cos^-1(1-y)) / lim as y→0 (y)
= 0/0 therfore limit of the entire function as x→1- is ∞

Last edited by a moderator: May 31, 2013
2. May 30, 2013

### SteamKing

Staff Emeritus
Sorry, 0/0 is an indeterminate form.

3. May 30, 2013

### mathgeek69

So 0/0 = Limit doesnt exist ?

4. May 30, 2013

### SteamKing

Staff Emeritus
Not necessarily.

For example, the limit of sin(x)/x as x approaches 0 is equal to 1.

5. May 31, 2013

### verty

Let $x = cos(y)$, this may help to simplify things.

If not, I would leave this question for later.

6. May 31, 2013

### Staff: Mentor

[0/0] is one of several indeterminate forms. The "indeterminate" part means that you can't tell if an expression with this form has a limit, and if it does, what that limit will be.

Some of the other indeterminate forms are [∞/∞], [∞ - ∞], and [1].

All of the limits below are of the [0/0] indeterminate form:
$$1. \lim_{x \to 0}\frac{x^2}{x}$$
$$2. \lim_{x \to 0}\frac{x}{x^2}$$
$$3. \lim_{x \to 0}\frac{x}{x}$$
In #1, the limit is 0; in #2, the limit doesn't exist; in #3, the limit is 1.

7. May 31, 2013

### eumyang

I think you switched #1 and #2. In #1, the limit doesn't exist, and in #2, the limit is 0.

8. May 31, 2013

### Staff: Mentor

I don't think so. Factor x/x out of each and see what you get.