Solving Limits with L'Hopital's Rule: Differentiating tan^-1(x-pi/4) for x->1

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In summary, L'Hopital's Rule is a mathematical theorem used in calculus to evaluate limits of indeterminate forms. It is used when the limit of a function approaches either 0/0 or infinity/infinity. The indeterminate form of tan^-1(x-pi/4) for x->1 is 0/0. To apply L'Hopital's Rule, take the derivative of the numerator and denominator separately and evaluate the limit again with the new simplified function. The derivative of tan^-1(x-pi/4) is 1/(1+(x-pi/4)^2) and the limit of the differentiated function is 0.
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JFonseka
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Homework Statement


lim
x->1 [tex]\frac{tan^{-1} (x - pi/4)}{x -1}[/tex]

Homework Equations


None that I know of


The Attempt at a Solution


Indeterminate form, so use L'Hopital's rule

Differentiate the top, then differentiate the bottom.
The differential of the denominator is just 1.
However I have no idea how to differentiate the numerator.

Thanks for any help
 
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  • #2
derivative of arctan(u) is: 1/(1+u^2)
 
Last edited:
  • #3
[tex]\frac{d}{dx}tan^{-1} (x) = \frac {1}{x^2+1}[/tex]
 
  • #4
Thanks guys.
 

What is L'Hopital's Rule and when is it used?

L'Hopital's Rule is a mathematical theorem used in calculus to evaluate limits of indeterminate forms. It is used when the limit of a function approaches either 0/0 or infinity/infinity.

What is the indeterminate form of tan^-1(x-pi/4) for x->1?

The indeterminate form of tan^-1(x-pi/4) for x->1 is 0/0.

How do you apply L'Hopital's Rule to solve this limit?

To apply L'Hopital's Rule, take the derivative of the numerator and denominator separately. Then, evaluate the limit again with the new simplified function.

What is the derivative of tan^-1(x-pi/4)?

The derivative of tan^-1(x-pi/4) is 1/(1+(x-pi/4)^2).

What is the limit of the differentiated function?

The limit of the differentiated function is 0.

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