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!