Problem with Limits using L'Hospital's Rule

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Homework Help Overview

The problem involves determining the limit of the expression lim ψ→1 of ψ^(1/(ψ-1)), which is related to the application of L'Hospital's Rule in the context of limits.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss different approaches to rewriting the limit expression, including the use of logarithms and L'Hospital's Rule. There is a comparison of two potential solutions, (a) and (b), both of which involve manipulating the logarithmic form of the limit.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the limit and the application of L'Hospital's Rule. Some guidance has been offered regarding rewriting the expression as a fraction before differentiation, but no consensus has been reached on the correct solution.

Contextual Notes

Participants are navigating the complexities of limits and logarithmic transformations, with some uncertainty about the correct approach and the implications of their manipulations.

CharlesL
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Homework Statement


Determine the limit of

lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 of [tex]\psi[/tex]^(1/([tex]\psi[/tex]-1))


Homework Equations


psigraph.jpg



The Attempt at a Solution



Let y = [tex]\psi[/tex]^(1/([tex]\psi[/tex]-1))
ln y = ln [tex]\psi[/tex]^(1/([tex]\psi[/tex]-1))

lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 ln y = lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 of (1/([tex]\psi[/tex]-1)) (ln [tex]\psi[/tex])

Differentiate

lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 ln y = -1/([tex]\psi[/tex]-1)2 x (1/[tex]\psi[/tex])

lim [tex]\psi[/tex][tex]\rightarrow[/tex]1 ln y = 2/([tex]\psi[/tex]3+3[tex]\psi[/tex]2+3[tex]\psi[/tex]+1)

ln y =1/4
y = e1/4

Does e1/4 = e?
 
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L'Hopital's rule works for f(x)/g(x) and then you get f'(x)/g'(x). Try rewriting the step before you differentiate as a fraction and not a product.
 
Thank you for your reply.

I wonder which is the correct solution

solution (a)

ln y = 1/([tex]\psi[/tex]-1) x ln [tex]\psi[/tex]

ln y = ln [tex]\psi[/tex] x ([tex]\psi[/tex]-1)

ln y = 1/[tex]\psi[/tex]

ln y = 1/1

y = e1

or solution (b)

ln y = 1/([tex]\psi[/tex]-1) x ln [tex]\psi[/tex]

ln y = ln [tex]\psi[/tex] / ([tex]\psi[/tex]-1)

ln y = 1/[tex]\psi[/tex]

ln y = 1/1

y = e1
 
Last edited:
I did it the second way, assuming you just didn't feel like typing out that you were still dealing with limits
 
Thank you Hogger for your point outs. Appreciate it. Have a nice day



Charles
 

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