# Help with limit

1. Sep 15, 2008

$$x\stackrel{lim}{\rightarrow}pi$$ sin(x) / (pi - x)

Sorry about the awkward looking notation, trying to understand how to use this stuff ^^

Anyways, any idea on how I can solve this? I am totally stumped. I know that sin(x) / x is one, but that damn pi symbol totally throws me :P

Thanks ^^

2. Sep 15, 2008

### tiny-tim

Welcome to PF!

(have a pi: π )

Two methods:

i] since you know the limit for sin(x)/x, just put y = π - x

ii] use the same method that you would have used to find sin(x)/x

(oh … and to get $$x\stackrel{lim}{\rightarrow}\pi \frac{sin x}{\pi - x}$$,

type [noparse]$$x\stackrel{lim}{\rightarrow}\pi \frac{sin x}{\pi - x}$$[/noparse])

3. Sep 15, 2008

### Feldoh

Well that's actually indeterminant, isn't it?

4. Sep 15, 2008

### HallsofIvy

Staff Emeritus
Since sin(x)/x is not "indeterminant" why would sin(x)/(pi- x) be? Just putting x= pi gives the "indeterminant" 0/0 which just means we can't do that to find the limit. There is nothing "indeterminant" about the limit.

adoado, you can also use sin(pi- x)= - sin(x) to rewrite sin(x)/(pi- x) as -sin(pi- x)/(pi-x). NOW replace "pi- x" by "y".

5. Sep 15, 2008

### Count Iblis

You don't need "\stackrel":

$$\lim_{x\rightarrow\pi} \frac{\sin(x)}{\pi - x}=1$$

6. Sep 15, 2008

### Feldoh

I meant that since it's in an indeterminant form, 0/0, l'hopital's rule comes into play.

7. Sep 15, 2008

### HallsofIvy

Staff Emeritus
That depends upon what you mean by "comes into play". In this case, the problem is much too simple to require L'Hopital's rule and I suspect that the OP has not yet had L'Hopital's rule.

8. Sep 15, 2008

### Feldoh

I mean you can use L'Hopital's rule to solve this. Either way is straight forward, but sometimes it's nice to know more than one way to solve a problem.

9. Sep 16, 2008