Finding the Limit of cos(pi*x / sinx)

[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?

 

[rock.freak667]

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

 

[rocomath]

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

Yes?

 

[sutupidmath]

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

Yes?

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?

 

[Oneiromancy]

Right.

Can I just use the quotient rule on the inside?

 

[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.

 

[Oneiromancy]

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

 

[sutupidmath]

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

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

 

[Oneiromancy]

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

 

[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?

 

[Oneiromancy]

Ya thanks it's -1.

 

[sutupidmath]

Ya thanks it's -1.

Good job!