# Finding right limit of log(2sin(x/2))-log(2(sin(x/2)+cos(x/2))

1. Aug 1, 2010

### Alexx1

log(2sin(x/2))-log(2(sin(x/2)+cos(x/2))

I have to find the right limit from x --> 0

How can I solve this limit?

Last edited: Aug 1, 2010
2. Aug 1, 2010

### Mentallic

Re: Limit

Have you tried anything? How about simplifying the two logarithms and then using double angle formulae on the trigs.

3. Aug 1, 2010

### Alexx1

Re: Limit

If you simplify them you get:

log(sin(x/2)/(sin(x/2)+cos(x/2))

But if I fill in 0 than I get log(0) but that doesn't exist..
If I use a double angle formulae I also get log(0)

4. Aug 1, 2010

### Mentallic

Re: Limit

Well that's why it's a limit which suggests you need to figure out what the limit is as x approaches 0 from the right side. Where is it heading?

5. Aug 1, 2010

### Alexx1

Re: Limit

If I plot the function on my calculator, I thinks it's heading to -8, but I'm not sure

6. Aug 1, 2010

### Mentallic

Re: Limit

That's because your calculator isn't accurate enough or you can't see exactly what's happening really close (even closer than that) to 0.

You've already figured out that the answer is ln(0+) - the plus is there to show it's from the right side, not both sides of 0. 0- denotes from the left side.
What is the answer to ln(0+)? That's the same as asking what happens to the function y=ln(x) as x approaches 0 from the right?

7. Aug 2, 2010

Re: Limit

Is it -∞?

8. Aug 2, 2010

### Mentallic

Re: Limit

Yep!

To convince yourself, try every smaller positive values of x,

x=0.1
x=0.001
x=10-10