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

[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?

 

[Mentallic]

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

 

[Alexx1]

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

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)

 

[Mentallic]

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?

 

[Alexx1]

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?

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

 

[Mentallic]

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?

 

[Alexx1]

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?

Is it -∞?

 

[Mentallic]

Yep!

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

x=0.1
x=0.001
x=10^-10

You'll see where it's headed.

 

[Alexx1]

Yep!

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

x=0.1
x=0.001
x=10^-10

You'll see where it's headed.

Thanks!