Solving the Limiting Issue: sin(x) / (π - x)

[adoado]

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 ^^
Adrian

 

[tiny-tim]

Welcome to PF!

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 ^^
Adrian

Hi Adrian! 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])

 

[Feldoh]

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

 

[HallsofIvy]

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

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".

 

[Count Iblis]

You don't need "\stackrel":


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

 

[Feldoh]

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".

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

 

[HallsofIvy]

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.

 

[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.

 

[adoado]

Hello ^^

Thanks everyone, I think I understand it now. And thanks for the syntax corrections :P

And about L'Hopitals Rule, indeed I have not learned it but its a good thing to know ;)

Thanks,
Adrian