Yet another limit question in this forum (No.2 :P)

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Homework Help Overview

The problem involves evaluating the limit of the difference of cosecant functions as t approaches x, specifically f(x) = lim t-> x [csc(t) - csc(x)]/(t - x), and finding the derivative at x = π/4.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the resemblance of the limit to the first principle definition of derivatives and explore various methods to approach the problem, including the application of L'Hospital's rule and trigonometric identities.

Discussion Status

Some participants have provided hints and suggestions, while others express uncertainty about specific identities and methods. There is an ongoing exploration of different approaches without a clear consensus on the best method to use.

Contextual Notes

One participant notes that they have not yet learned L'Hospital's rule, indicating a potential constraint in their approach. Additionally, there is mention of the problem being in an indeterminate form, which is a key aspect of the discussion.

Ballox
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Homework Statement


f(x)= lim t-> x [csc(t)-csc(x)]/(t-x). Find the value of f'(PI/4)


Homework Equations


I can see that this equation somewhat resembles one of the first principle def'ns

lim z->x [f(z)-f(x)]/(z-x)


The Attempt at a Solution


Not really sure how to begin.
I converted the csc(t) and the csc(x) to 1/sin(t) and 1/sin(x) and did a common denominator there, but I'm not sure if that's the correct approach to solving this problem.

I'm open to any suggestions and thank you for your time
Ballox
 
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Hi Ballox! :smile:

(have a pi: π :wink:)

Hint: one of the standard trigonometric identities is sin(t) - sin(x) = 2cos((t+x)/2)sin((t-x)/2) :wink:
 
Last edited:
I don't recognize that as a standard trigonometric identity :(
 
Since it's of the indeterminate form 0/0, applying L'Hospital's rule should help :wink:
 
Mentallic said:
Since it's of the indeterminate form 0/0, applying L'Hospital's rule should help :wink:

Haven't learned L'hopital's rule yet.
I believe we're supposed to use other methods
 
(just got up :zzz: …)
Ballox said:
I don't recognize that as a standard trigonometric identity :(

Then you will next time! :smile:

You can check for yourself that it's correct

(and familiarise yourself with the similar ones in the PF Library on trigonometric identities :wink:)
 
Ballox said:

Homework Statement


f(x)= lim t-> x [csc(t)-csc(x)]/(t-x). Find the value of f'(PI/4)


Homework Equations


I can see that this equation somewhat resembles one of the first principle def'ns

lim z->x [f(z)-f(x)]/(z-x)
Quite correct- that is the "first principle" definition of the derivative of csc(x). So your problem is really "evaluate the second derivative of csc(x) at x= \pi/4".


The Attempt at a Solution


Not really sure how to begin.
I converted the csc(t) and the csc(x) to 1/sin(t) and 1/sin(x) and did a common denominator there, but I'm not sure if that's the correct approach to solving this problem.

I'm open to any suggestions and thank you for your time
Ballox
 
Hmmm. So I read through all your responses and would like to thank you all for your help.

I looked back at the question and this is my solution:

limt->x [csc(t)-csc(x)]/(t-x) = d/dx csc(x) (not sure if this is represented correctly, but I see some sort of relationship here)

=> f(x)= -csc(x)cot(x)
f'(x)= -csc(x)*-csc^2(x) + cot(x)(csc(x)cot(x))
f'(x)= csc^3(x) + cot^2(x)(csc(x))

f'(PI/4)= 1/(sin^3(PI/4)) + [1/(tan^2(PI/4)) * 1/(sin(PI/4))]
f'(PI/4)= 2SQRT2 + (1*SQRT2)
f'(PI/4)= 2SQRT2 + SQRT2
f'(PI/4) = 3SQRT 2

Would this be correct?
 
Ballox said:
f'(PI/4) = 3SQRT 2

Would this be correct?

Yes. :smile:
 

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