# Limits of log sin equal to 0/1/infinity?

1. Jul 28, 2011

### LASmith

1. The problem statement, all variables and given/known data

limx->$\pi$/2(ln sin x)/(($\pi$-2x)2)

2. Relevant equations

3. The attempt at a solution

Putting $\pi/2$ into this equation gives 0/0, however, the limit of this could be 1,0 or infinity, so how would you distinguish between the three?

2. Jul 28, 2011

### tiny-tim

Hi LASmith!

l'Hôpital's rule?

3. Jul 28, 2011

### uart

LASmith, note that it doesn't have to be one of those three options.

4. Jul 28, 2011

### Staff: Mentor

This being precalculus, I think it would be sufficient to evaluate the function for a few values very close to Pi/2, on one side then the other. For example, Pi/2 - 0.001, Pi/2 - 0.002, Pi/2 - 0.005 and also try Pi/2 + 0.001 etc. (Then graph the points, at least in your mind, so you can see what the function is doing in the vicinity of Pi/2.)

Ideally, you might think it would be useful to evaluate Pi/2 - 0.00000000001 etc. so as to be really close to the point of interest, Pi/2, but this many sig figs will exceed the capabilities of your calculator. There is software which will implement a calculator that can work accurately to hundreds of decimal places, but we really don't need such extravagance here. (Though it is fun to experiment with to confirm what happens really really close to difficult points. )

5. Jul 29, 2011

### PeterO

That is why we have limx->$\pi$/2

We consider what happens when x gets very close to $\pi$/2 [from above and below] but never actually equaling $\pi$/2.

6. Jul 29, 2011

### gat0man

Consider: What are both of the functions doing as X-> pi/2?

As you are in precalc and not expected to use L'Hopitals Rule, look at the graphs of these functions and consider which has larger values as X-> pi/2

If you have a very small value in the numerator and a larger value in the denominator, what happens?