Limit Evaluation

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



I had an exam last night and I was stuck on this problem.
I went home where I had more time (and resources) and still can't figure it out, and I won't get my exam back (it was a final).

[itex]\lim_{x\to0}\frac{1}{x^{2}}-\frac{1}{xsinx}[/itex]

Homework Equations





The Attempt at a Solution


I have tried combining into one fraction, splitting up in to two separate limits, I can't use L'Hopital's as it is 1/inf.
What am I missing? The rest of the exam was somewhat easy, but I didn't even know where to start!
Can someone help me work though it? I think I just need help with a first step.
Thanks!
 

Answers and Replies

  • #2
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If you combine them into one fraction, then it's 0/0 right? Only problem is that you have to use L'Hopital 3 times to get the result.
 
  • #3
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Hello crybllrd write your expression in the following form:

[tex]\lim_{x\rightarrow0}-\frac{x^{2}sin(x)-xsin(x)^{2}}{x^{3}sin(x)^{2}}[/tex]

I think it should be clear from here on.
 
  • #4
CompuChip
Science Advisor
Homework Helper
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You can combine ito one fraction and then apply L'Hopitals repeatedly.

Or you can do a series expansion.
 
  • #5
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Hello crybllrd write your expression in the following form:

[tex]\lim_{x\rightarrow0}-\frac{x^{2}sin(x)-xsin(x)^{2}}{x^{3}sin(x)^{2}}[/tex]

I think it should be clear from here on.
How did you get the sin(x)2 in the denominator? If I multiply x2by xsin(x) I get x3sin(x).
 
  • #6
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If you combine them into one fraction, then it's 0/0 right? Only problem is that you have to use L'Hopital 3 times to get the result.
I used l'hopital's rule 6 times, not 3 times.
 
  • #7
HallsofIvy
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Homework Helper
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How did you get the sin(x)2 in the denominator? If I multiply x2by xsin(x) I get x3sin(x).
Why would you do that? The "least common denominator" is [itex]x^2sin(x)[/itex]. Of course, you also do not get "[itex]x^2sin^2(x)[/tex]"!

[tex]\frac{1}{x^2}- \frac{1}{x sin(x)}= \frac{sin(x)}{x^2 sin(x)}- \frac{x}{x^2 sin(x)}[/tex]
[tex]= \frac{sin(x)- x^2}{x^2 sin(x)}[/tex]

Now, you can use L'Hopital's rule on that, just once. Differentiating both numerator and denomimator separately you get
[tex]\frac{cos(x)- 2x}{2x sin(x)+ x^2cos(x)}[/tex].

You do NOT get "1/infinity".
 
  • #8
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Ah, ok, I got -1/6.
I don't remember why exactly I didn't get a good answer on the exam, but I think I was on like my 6th L'Hopital iteration and wasn't willing to spend any more time on a 4 point problem.
Thanks guys, now I can rest easy this summer, as my brain is completely fried now!
 
  • #9
1,225
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Hi. As already other colleagues have stated,

[itex]\lim_{x\to0}[\frac{1}{x^{2}}-\frac{1}{xsinx}][/itex]

[itex]=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - \frac{x}{sinx}][/itex]

[itex]=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - \frac{1}{1 - \frac{x^{2}}{3!} + ...} ][/itex]

[itex]=\lim_{x\to0}\frac{1}{x^{2}}[ 1 - 1 - \frac{x^{2}}{3!} + ...][/itex]
 

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