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

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In summary, the conversation discusses the limit of the function sin(x)/(pi-x) as x approaches pi. Two methods are suggested to solve this problem, including using the known limit of sin(x)/x and using the trigonometric identity sin(pi-x)=-sin(x). The use of L'Hopital's rule is also mentioned, although the OP has not yet learned it.
  • #1
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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  • #2
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])
 
  • #3
Well that's actually indeterminant, isn't it?
 
  • #4
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".
 
  • #5
You don't need "\stackrel":


[tex]\lim_{x\rightarrow\pi} \frac{\sin(x)}{\pi - x}=1[/tex]
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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
 

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

What is the limit of sin(x) / (π - x) as x approaches π?

The limit of sin(x) / (π - x) as x approaches π is undefined. This is because the denominator becomes 0, which results in an indeterminate form.

How can I solve the limiting issue of sin(x) / (π - x)?

The limiting issue of sin(x) / (π - x) can be solved by using L'Hôpital's rule. This rule states that if the limit of a function is in an indeterminate form, then the limit of the ratio of the derivatives of the numerator and denominator will be the same as the original limit.

Can I use any other methods to solve the limiting issue of sin(x) / (π - x)?

Yes, there are other methods that can be used to solve this limiting issue, such as substitution and algebraic manipulation. However, L'Hôpital's rule is often the most efficient and straightforward method.

What is the significance of the limiting issue of sin(x) / (π - x)?

The limiting issue of sin(x) / (π - x) is significant because it helps us understand the behavior of the function as x approaches π. It also allows us to find the value of the limit, which can be useful in various mathematical and scientific applications.

Are there any real-life applications of solving the limiting issue of sin(x) / (π - x)?

Yes, the limiting issue of sin(x) / (π - x) can be applied in various fields such as physics, engineering, and economics. For example, in physics, it can be used to calculate the maximum tension in a string under tension, and in economics, it can be used to determine the maximum profit a company can make with limited resources.

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