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Problem with Limits using L'Hospital's Rule

  • Thread starter CharlesL
  • Start date
1. Homework Statement
Determine the limit of

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


2. Homework Equations
psigraph.jpg



3. 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?
 
21
0
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:
21
0
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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