Find the limit of ln (sinx) as x approaches pi

[essedbl]

I need to find the limit of. Ln (sinx) as x approaches pi-, that is, as x approaches pi from the left.

I thought I should use the squeeze theorem, but I am not sure how to apply it. My teacher is not requiring us to use the delta/Epsilon method, so I am sure it is a squeeze method. But how do I set it up?

I tried -1 < sinx < 1,
Thus ln (-1) < ln (sinx) < ln (1)

But you can't take the ln of a negative... please help.

 

[jbunniii]

What can you say about \sin(x) as x \rightarrow \pi^-? Is it positive or negative? What value does it approach?

 

[Sayajin]

As x aproaches pi from the left your sine function aproaches 0. This means this is equivalent of finding the limit as the thing inside the natural log aproaches 0. Its very easy limit.

 

[essedbl]

Oh. So this isn't squeeze action... I just find the limit of ln as sinx approaches zero. Hmm... I would need to graph that. But that should be simple enough. Tjanks guys

 

[Mark44]

Oh. So this isn't squeeze action... I just find the limit of ln as sinx approaches zero. Hmm... I would need to graph that.

Good idea. Remember that x is approaching π from the left, which says something about how sin(x) is approaching zero, and finally, what ln(sin(x)) is doing.

But that should be simple enough. Tjanks guys