L'Hopital's Rule

  • Thread starter chlrbwls
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  • #1
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So I've been working on my analysis hw and im stuck on the last question!

Please help!

the question is

lim x approaches to 0 of (x/sin x)^1/x^2

the answer is e^1/6

but my l'Hopital's process keeps giving me 0/0...

After taking the log of the limit, I have

====> 1/x^2 (log x - log sin x)

My first l'Hopitals process give me

====> 1/2x (1/x - cos x/ sin x)

2nd l'Hopitals give me

====> 1/2 (-(x)^-2 - 1/sin^2 x)

Am i doing something wrong? or do i need to keep going
 

Answers and Replies

  • #2
tiny-tim
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hi chlrbwls! :smile:

(try using the X2 icon just above the Reply box :wink:)
====> 1/2 (-(x)^-2 - 1/sin^2 x)

good so far :smile: (apart from a sign) …

now write the bottom as (x2 - sin2x)/x2sin2x, and expand :wink:
 
  • #3
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What do you mean by expand tim?

Once i expand, do i need to do l'Hopitals again?
 
  • #4
tiny-tim
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do you know an expansion (a series) for sinx ?

(and no, you wouldn't need l'Hôpital again)
 
  • #5
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Oh you mean the taylor polynomial series!

But i have squares in with the sines..

How would that work?
 
  • #6
tiny-tim
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well, you'll only need the first two terms, so forget all the other terms, and square it! :smile:

(alternatively, use 2sin2x = 1 - cos2x :wink:)
 
  • #7
jambaugh
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I haven't looked at the detail but when using l'Hospital's rule be sure to check each time that you do indeed have the indeterminate form necessary for the rule to be valid.

E.g. naively over-applying l'Hospital's rule yields:

x/x -> 1/1 -> 0/0 (the 1/1 form is not indeterminate so d1/dx / d1/dx = 0/0 is incorrect.)
 
  • #8
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Yeah the only indeterminate forms are 0/0 and infinity/infinity. So unless the function yields either of these when the limit is plugged in, it could be solved algebraically.
 
  • #10
uart
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hi chlrbwls! :smile:

(try using the X2 icon just above the Reply box :wink:)


good so far :smile: (apart from a sign) …

now write the bottom as (x2 - sin2x)/x2sin2x, and expand :wink:

Tim there's no need to apply L'hopitals twice before applying series expansion. It's much easier to do that right after the first application.

====> 1/2x (1/x - cos x/ sin x)

(sinx - xcosx) / (2 x^2 sinx)

(-1/6 x^3 + 1/2 x^3) / (2 x^3 ) = 1/6
 

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