# Difficulty to find this limit of trigonometric function

1. Apr 23, 2012

1. The problem statement, all variables and given/known data
Hello! I'm having some trouble to find this limit $$\lim_{x\rightarrow \pi}\frac{senx}{x-\pi }$$ Also, limits like this one $$\lim_{x\rightarrow \frac{\pi}2{}}\frac{1-senx}{2x-\pi }$$ confuse me a lot!

2. Relevant equations

3. The attempt at a solution
I don't even know where to start! I tried this but it's problably just waste of time.
My idea was try calculate the area of the third sector of the trigonometric circle. Because to me that -∏ is making my area goes to the third sector, the negative one. Using this idea I've done:
$$\frac{-senx}{senx}< \frac{-x}{senx}<\frac{-tgx}{senx}$$
Dividing by senx we got
$$-1> \frac{senx}{x}>-1$$
But I'm stuck here. Help please!

And forgive me for my bad english! English is not my native language.

Last edited: Apr 23, 2012
2. Apr 23, 2012

### Staff: Mentor

You can rewrite this as
$$\lim_{x \to \pi}\frac{sin(\pi - x)}{x - \pi}$$

Then, let u = x - $\pi$.

There's a very common limit that will come in handy here.

3. Apr 23, 2012

### Staff: Mentor

BTW, in English we call it sine (or sin).

4. Apr 23, 2012

Thanks for your answer but I still don't get it! Why can I rewrite my limit as?
$$\lim_{x\rightarrow \pi } \frac{sin(\pi -x)}{x-\pi}$$
Well, assuming that it's ok to rewrite the limit as above I think I know what to do next:
$$\lim_{x\rightarrow \pi } \frac{sin(-(\pi -x))}{x-\pi}$$
$$\lim_{x\rightarrow \pi } \frac{-sin(x-\pi)}{x-\pi}$$
$$\lim_{x\rightarrow \pi } \frac{-sin(x-\pi)}{x-\pi}=-1$$

5. Apr 23, 2012

### Staff: Mentor

1. Because sin(x) = sin($\pi$ - x)
2. If u = x - $\pi$, then as x approaches $\pi$, u approaches 0. Your limit is $\lim_{u \to 0} \frac{-sin(u)}{u}.$

6. Apr 23, 2012

### Staff: Mentor

For the other problem, a similar approach will work, I believe.

7. Apr 23, 2012

I just didn't get why sin(x) = sin(π - x). I tried to solve this one by doing what you did $$\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sinx}{2x-\pi }$$
this is result
$$\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sinx}{2x-\pi }=\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-sin(\pi -x)}{2x-\pi }=\lim_{x\rightarrow \frac{\pi }{2}}\frac{1+sin(x-\pi )}{2x-\pi }\frac{2}{2}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{2+2sin(x-\pi)}{4x-2\pi}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{2+2sin(x-\pi)}{2(2x-\pi)}=\lim_{x\rightarrow \frac{\pi }{2}}\frac{2}{2}-\frac{1}{1} = 0$$
This is totally wrong :grumpy:

8. Apr 23, 2012

### Staff: Mentor

sin(x) = cos($\pi$/2 - x)

2x - $\pi$ = 2(x - $\pi$/2)

9. Apr 23, 2012

I still don't know how to solve $$\lim_{x\rightarrow \frac{\pi}2{}}\frac{1-senx}{2x-\pi }$$

10. Apr 23, 2012

### Staff: Mentor

Post #8 has some clues. Do you know of any special trig limits?

11. Apr 23, 2012

$$\lim_{x\rightarrow \frac{\pi }{2}}\frac{1-cos(x-\frac{\pi }{2})}{2(x-\frac{\pi }{2})} = \lim_{x\rightarrow \frac{\pi }{2}} \frac{1-1}{2}=0$$

12. Apr 23, 2012

### Staff: Mentor

You have a mistake. The denominator is approaching 0, not 2.

The other limit I referred to is
$$\lim_{t \to 0}\frac{1 - cos(t)}{t} = 0$$