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

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Discussion Overview

The discussion revolves around finding the limit of the expression sin(x) / (π - x) as x approaches π. Participants explore various methods to solve this limit, including the application of known limits and L'Hôpital's rule. The conversation includes technical reasoning and clarifications regarding the nature of indeterminate forms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Adrian expresses confusion about solving the limit of sin(x) / (π - x) as x approaches π, noting familiarity with the limit of sin(x) / x.
  • One participant suggests substituting y = π - x to simplify the limit calculation.
  • Another participant points out that the limit results in an indeterminate form (0/0) when directly substituting x = π, questioning the characterization of the limit as indeterminate.
  • There is a discussion about using the identity sin(π - x) = -sin(x) to rewrite the expression for further analysis.
  • Some participants argue that L'Hôpital's rule is unnecessary for this particular limit, while others suggest it as a valid method to solve the problem.
  • Adrian acknowledges the help received and expresses understanding of the limit, as well as appreciation for the syntax corrections provided by participants.

Areas of Agreement / Disagreement

Participants have differing views on the necessity of L'Hôpital's rule for this limit problem. While some believe it is too simple to require such a method, others advocate for knowing multiple approaches. The discussion remains unresolved regarding the characterization of the limit as indeterminate.

Contextual Notes

The discussion includes varying assumptions about the participants' familiarity with L'Hôpital's rule and the concept of indeterminate forms. There is also a reliance on specific mathematical identities and limit properties that may not be universally understood.

adoado
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[tex]x\stackrel{lim}{\rightarrow}pi[/tex] 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
 
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Welcome to PF!

adoado said:
[tex]x\stackrel{lim}{\rightarrow}pi[/tex] 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! :smile:

(have a pi: π :smile:)

Two methods:

i] since you know the limit for sin(x)/x, just put y = π - x :wink:

ii] use the same method that you would have used to find sin(x)/x :smile:

(oh … and to get [tex]x\stackrel{lim}{\rightarrow}\pi \frac{sin x}{\pi - x}[/tex],

type [noparse][tex]x\stackrel{lim}{\rightarrow}\pi \frac{sin x}{\pi - x}[/tex][/noparse])
 
Well that's actually indeterminant, isn't it?
 
Feldoh said:
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".
 
You don't need "\stackrel":


[tex]\lim_{x\rightarrow\pi} \frac{\sin(x)}{\pi - x}=1[/tex]
 
HallsofIvy said:
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.
 
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.
 
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.
 
Hello ^^

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

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

Thanks,
Adrian
 

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