L'Hopital's rule

  • Thread starter RockPaper
  • Start date
  • #1
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Homework Statement


lim as x -> 0 sin(e^(x^2)-1)/e^(cosx)-e
See Link Below.

I saw this on the net and tried to solve it. I'm wondering if I'm correct by any means or did I mess up somewhere along the line.

Homework Equations




The Attempt at a Solution


http://img291.imageshack.us/img291/1174/hopitalsrule.jpg [Broken]
 
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Answers and Replies

  • #2
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The second differentiation of the numerator isn't correct. It shouldn't be just one term but two.
(f*g)' = fg' + gf'
 
  • #3
216
1
You did it well.

Regards.
 
  • #4
5
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The second differentiation of the numerator isn't correct. It shouldn't be just one term but two.
(f*g)' = fg' + gf'

You did it well.

Regards.

I think you're right Bohrok (because I have a tendency of messing up).

@Bohrok: Do you mean the second differentiation should be

-sin(ex2-1)2xex2+(2ex2+4x2ex2)cos(ex2-1)/-cos(x)ecosxsin2x*ecosx

also, would the negative signs go away since we have both in the numerator and denominator? could we reduce the answer to 0 or do we keep it as 0/e?
 
  • #5
35,236
7,056
I agree with Bohrok and disagree with njama. Let's take it a step at a time. Your first differentiation of the numerator was correct, but the second one wasn't, and your latest try doesn't look right either.

What do you get for
[tex]\frac{d}{dx} 2xe^{x^2}~cos(e^{x^2} - 1)[/tex] ?

Also, you should NOT leave an answer as 0/e.
 
  • #6
5
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I agree with Bohrok and disagree with njama. Let's take it a step at a time. Your first differentiation of the numerator was correct, but the second one wasn't, and your latest try doesn't look right either.

What do you get for
[tex]\frac{d}{dx} 2xe^{x^2}~cos(e^{x^2} - 1)[/tex] ?

Also, you should NOT leave an answer as 0/e.

Thank you for clearing that up, so the limit as x->0 is 0

I get:
2ex2+4x2ex2cos(ex2-1)-sin(ex2-1)*2xex2*2xex2
 
  • #7
35,236
7,056
Thank you for clearing that up, so the limit as x->0 is 0

I get:
2ex2+4x2ex2cos(ex2-1)-sin(ex2-1)*2xex2*2xex2

Close, but you are missing a pair of necessary parentheses and one exponent is incorrect, according to my work.
Edit: the 3 exponent below should be 2.
(2ex2+4x3ex2)cos(ex2-1)-sin(ex2-1)*2xex2*2xex2

BTW
[tex]2xe^{x^2}*2xe^{x^2} = 4x^2e^{2x^2}[/tex]
 
Last edited:
  • #8
5
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Close, but you are missing a pair of necessary parentheses and one exponent is incorrect, according to my work.

(2ex2+4x3ex2)cos(ex2-1)-sin(ex2-1)*2xex2*2xex2

BTW
[tex]2xe^{x^2}*2xe^{x^2} = 4x^2e^{2x^2}[/tex]

I'm glad I finally got it because it took me a few tries to get it :/ which is typical.
I don't understand how it's 4x3.
And thank you for telling me that it's 4x^2e^(2x^2) because I didn't know how to simplify it :shy:
 
  • #9
35,236
7,056
You're right - that exponent shouldn't be 3. It took me several times checking to find what I did wrong.

[tex]d/dx(2xe^{x^2}~cos(e^{x^2} - 1)) = 2xe^{x^2}(-sin(e^{x^2} - 1) \cdot 2xe^{x^2}) + d/dx(2xe^{x^2}) \cdot cos(e^{x^2} - 1)[/tex]
[tex]=-4x^2e^{2x^2}sin(e^{x^2} - 1) + (2e^{x^2} + 2x \cdot 2xe^{x^2})cos(e^{x^2} - 1)[/tex]
[tex]= (2e^{x^2} + 4x^2e^{x^2})cos(e^{x^2} - 1) - 4x^2e^{2x^2} sin(e^{x^2} - 1)[/tex]
 
  • #10
5
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Great! Thank you for the help... I actually added the parenthesis on my paper but somehow didn't do that when I wrote it on the computer. :| It took me a whole bunch of tries to get this right (even when I had an idea on what to do), I need to work on my math skills which I'm hoping to get better over the summer. Hopefully I'll finally gain full understanding on math once I practice and learn the rules.
 

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