# Find the limit.

1. Feb 5, 2008

### Oneiromancy

Find the limit.

limit x -> 0 cos(pi*x / sinx)

I'm only in cal I. I was hoping I wouldn't have to use the difference quotient so is there an easier way to do this using derivative rules or something?

2. Feb 5, 2008

### rock.freak667

$$lim_{x\rightarrow 0} cosf(x) =cos (lim_{x\rightarrow 0} f(x))$$

3. Feb 5, 2008

### rocomath

$$\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}$$

Yes?

4. Feb 5, 2008

### sutupidmath

Yeah this is what he meant, i guess. And follow rock.freak667's advice. But remember that this is true only because cos is continuous for any real.
$$lim_{x\rightarrow 0} cosf(x) =cos (lim_{x\rightarrow 0} f(x))$$
Now can you see the trick after you go inside the cosine function with the limit?

5. Feb 5, 2008

### Oneiromancy

Right.

Can I just use the quotient rule on the inside?

6. Feb 5, 2008

### sutupidmath

well you do not need to use the quotient rule, because do you know what the limit of

sin(x)/x is as x--->0

lim(x-->0)(sin(x))/x ----?????
Just use this fact and you will be fine, because if you use the quotient rule as you are claimint to, you will get an intermediate form of 0/0.

7. Feb 5, 2008

### Oneiromancy

This would be x / sin x in this case. You got it backwards.

8. Feb 5, 2008

### sutupidmath

NO, i did not get it backwards, because if you know what lim(x-->0)(sin(x))/x ----?????
is you will not have problem finding out what
lim(x-->0)x/sinx is, since you will take its reciprocal and everything will turn into terms of sin(x)/x

9. Feb 5, 2008

### Oneiromancy

Oh, sorry, calm down lol. It's 1.

10. Feb 5, 2008

### sutupidmath

$$\lim_{x\rightarrow 0}\cos{\left(\frac{\pi x}{\sin x}\right)}=\cos{(\lim_{x\rightarrow 0}\left\frac{\pi}{\frac{\sin x}{\ x}\right)}}$$
do u see now what to do?

11. Feb 5, 2008

### Oneiromancy

Ya thanks it's -1.

12. Feb 5, 2008

Good job!