Limit problem

  • Thread starter lab-rat
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  • #1
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Homework Statement



Evaluate limx->0 (e^x - 1- x - (x^2/2))/x^3

The Attempt at a Solution



I can't remember how to solve this limit. Do I need to evaluate each part seperately? I plugged in the 0 to find that the limit does exist. I just can't seem to figure out what to do next.
 

Answers and Replies

  • #2
353
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Why not use l'Hôpital's rule?
 
  • #3
22,129
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Or use the Taylor series of the exponential function.
 
  • #4
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If I use l'Hopital's rule won't I end up with 3x^2 at the bottom?
 
  • #5
22,129
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You will want to use l'Hopital multiple times.
 
  • #6
The limit of a sum is the sum of limits - Lim[a+b]=Lim[a]+Lim
Then as everyone before has said, you're going to want to use l'hopitals rule for the last term
 
  • #7
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I used L'Hopital's rule three times and ended up with limx->0(e^x)/5

Is this correct? It still gives me 0 on top.
 
  • #8
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Here is how I did it in case it might help

lim->0 (e^x)' - 1' - x' - (x^2/2)' /x^3'

= lim->0 (e^x' - 1' - x')/ 3x^2'

= lim->0 (e^x' - 1') / 5x'

= lim x->0 (e^x) /5
 
  • #9
353
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Very close, but look at your second to thrid line in the denominator.

What is: $$\frac{d}{dx}(3x^{2})$$

After you fix that, evaluate it at 0.

[tex]
\lim_{x\to 0}~ e^{x}
[/tex] should not be 0.
 
Last edited:
  • #10
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Thanks!
It should be 6x right?

So is 1/6 the correct answer?
 
  • #11
SammyS
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1/6 is the correct result.
 

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