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Limit as x approaches 0.

  • #1

Homework Statement


http://www.prep101.com/files/Math100PracticeExam.pdf [Broken]

Question 1m

The Attempt at a Solution


I tried to do this by using
lim sinx/x = 1
x->0

Factoring out x^2 from each one and I get infinity.
Why does this method not work?

The answer says it's 4/5.

I even tried plugging in 0.001 in my calculator and I get infinity :S.
 
Last edited by a moderator:

Answers and Replies

  • #2
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Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?
 
  • #3
Dick
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Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?
Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.
 
  • #4
Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.
It is 3sin(2x^2), you get 4/5, this limit is INSANE :P
 
  • #5
vela
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It helps to use the substitution u=x2 to get
[tex]\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}[/tex]Then three applications of the Hospital rule gets you to a limit you can evaluate.
 
  • #6
Dick
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It helps to use the substitution u=x2 to get
[tex]\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}[/tex]Then three applications of the Hospital rule gets you to a limit you can evaluate.
Ah, indeed it does.
 

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