# Trigonometric limit question

1. Oct 13, 2008

### shocker121

1. The problem statement, all variables and given/known data
Lim (1-cosX)/X^2
X->0

2. Relevant equations
Not 100% sure try to use the following identity:
sin^2(X)=(1-cosX)(1+cosX) or
sin^2(X)=1-cos^2(X)

3. The attempt at a solution
Tried substituting (sin^2(X))/(1+cosX)) for 1-cosX but that didnt help and so for the last half hour i've been staring at this problem and I am still lost. Somebody help me out.
Possible answers were 0, .5, 1, 2

2. Oct 13, 2008

### Staff: Mentor

There are at least a couple of ways to do this, but one of them might not be applicable based on which class you're in.

Try multiplying the numerator and denominator by (1 + cos(x))/(1 + cos(x)). Then make your substitution. If you do that, the limit, as x --> 0, will be
lim $$\frac{sin^2(x)}{x^2(1 + cos(x))}$$
Now, do you know any limits that involve sin(x) and x, as x --> 0?

3. Oct 13, 2008

### shocker121

Im in AP calculus B/C if it affects your answer at all
and if i'm not mistaken

limit sin(x)/x = 1
x-> 0

and

limit sin^2(x)/x^2 = 1
x-> 0

but how would i go about seperating the the 1+cos(x) from the denominator

could i make the function

limit (sin^2(x)/x^2)*(1/(1+cosx))
x-> 0

then substitute in for x?

Last edited: Oct 13, 2008
4. Oct 13, 2008

### Staff: Mentor

Not quite, but a property of limits is that lim (A * B) = lim A * lim B, providing that all the limits exist. Is that enough of a hint?

The other approach is to look at the infinite series for 1 - cos(x). I hesitated to point you in that direction since I didn't know where you were in your course. If you're interested in the details of this approach, let me know.

Mark

5. Oct 14, 2008

### HallsofIvy

Staff Emeritus
Yes, you certainly can write
$$\frac{1- cos(x)}{x^2}= \frac{(1- cos(x))(1+ cos(x))}{x^2(1+ cos(x))}$$

$$= \frac{1-cos^2(x)}{x^2(1+ cos(x))}= \frac{sin^2(x)}{x^2(1+ cos(x))}$$

$$= \left(\frac{sin(x)}{x}\right)^2 \frac{1}{1+ cos(x)}$$

That should make the limit easy.