Limit, and lopitals rule

  • Thread starter gayani
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  • #1
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Homework Statement



Find A so that lim (x+A/x-2A)^x =5
x-> 00

Homework Equations





The Attempt at a Solution


gotta use lopihtals rule.
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Hi gayani! Welcome to PF! :smile:

(have an infinity: ∞ and try using the X2 tag just above the Reply box :wink:)
Find A so that lim (x+A/x-2A)^x =5
x-> 00

gotta use lopihtals rule.
hmm … I can't see how to use l'Hôpital's rule :redface:

but if you write the LHS as (1 + 3A/(x - 2A))x, and fiddle about with it, you should get something you recognise. :smile:
 
  • #3
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Gayani,
Just in case you don't follow what Tiny-tim did, here it is.

You wrote the rational expression as x + A/x - 2A. What I'm pretty sure you meant (and the way Tiny-tim interpreted it) was (x + A)/(x - 2A). What you wrote would be correctly interpreted as the sum and difference of three expressions-- x, A/x, and 2/A, not the quotient of x + A and x - 2A.

What Tiny-tim did was divide x + A by x - 2A, so that (x + A)/(x - 2A) = 1 + 3A/(x - 2A).
 
  • #4
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Take the natural log of both sides and then take the limit of the natual log of the left side

[tex]\lim_{x\rightarrow\infty} ln\left(1 + \frac{3A} {x-2A}\right)^x= ln (5)[/tex]

You'll have to do some rearranging in order to use l'Hôpital's rule and "solve" for A.
 
Last edited:

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