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Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible?)

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data


    Calculate the following limit:

    [tex]\lim_{x\rightarrow0}\frac{1}{x^{2}}-\frac{1}{xsinx}[/tex]




    3. The attempt at a solution

    I've tried everything possible and i keep getting undefined forms like 0/0 and ∞-∞. Any ideas anyone??
     
    Last edited: May 13, 2012
  2. jcsd
  3. May 13, 2012 #2

    sharks

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    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    It is indeed an indeterminate form and you should use L'Hopital's Rule:
    [tex]\lim_{x\rightarrow0}\frac{1}{x^{2}}-\frac{1}{xsinx}=\lim_{x\rightarrow0}\frac{\sin x-x}{x^2 \sin x}=\frac{0}{0}[/tex]
     
  4. May 13, 2012 #3

    SammyS

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    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    Can you list what you've tried in more detail so we don't go around in circles?
     
  5. May 13, 2012 #4
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    I've tried getting a common denominator like sharks stated, and then i tried using trig rules which i couldn't and then now i'm stuck.
     
  6. May 13, 2012 #5
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    I was able to solve it easily using l'hopital's rule 3 times repeatedly and i got an answer of -1/6, but doing it without l'hopital's rule is really difficult.
     
  7. May 13, 2012 #6
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    Using the Taylor series expansion for sin(x) isn't difficult...if you have an indeterminate form limit, trying to algebraically shuffle your way out of it won't help you very much.
     
  8. May 13, 2012 #7

    sharks

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    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    3 times? It should be only once.
     
  9. May 13, 2012 #8
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    No, It took me 3 times, because I didn't use taylor expansion.
     
  10. May 13, 2012 #9

    sharks

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    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    [tex]\lim_{x\rightarrow0}\frac{\sin x-x}{x^2 \sin x}=\lim_{x\rightarrow0}\frac{\cos x-1}{2x \sin x+ x^2\cos x}=\frac{0}{0}[/tex]
     
  11. May 13, 2012 #10
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    ... which is indeterminate.
     
  12. May 13, 2012 #11

    sharks

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    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    The answer is: [itex]-\frac{1}{6}[/itex]. You are correct. But i don't see any other method.
     
  13. May 13, 2012 #12
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    Well, you know that [itex]\displaystyle\lim_{x\to 0}\frac{1}{x^2}-\frac{1}{x\sin x} = -\frac{1}{6}[/itex] (using l'hopital's rule), but, you want to prove it without using l'hopital's rule, so you prove it by definition ((ε,δ)-definition of limit)

    so, you use l'hopital's rule to see what is the limit, and then, you prove it by definition (without l'hopital's rule)

    sorry for bad English
     
  14. May 13, 2012 #13
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    Does the problem ask you to solve the limit without taylor series or l'Hospital's rule, or are you just wondering if it can be solved without those methods?
     
  15. May 14, 2012 #14
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    No i'm just wondering if it could be solved without them.
     
  16. May 14, 2012 #15
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    I don't want to prove the limit i just want to solve it.
     
  17. May 14, 2012 #16
    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    And then derive what you have 2 more times and you'll get -cos(x)/something really long.

    Then when you compute it you will get -1/6, but I don't get how you did it using l'hopitals rule once only??
     
  18. May 14, 2012 #17

    Curious3141

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    Re: Solving a hard limit without taylor's series nor l'hopitals rule ( is it possible

    The easiest way is to first bring everything to a common denominator, then use the Taylor series for sin(x) to the [itex]x^3[/itex] term on both numerator and denominator. Simplify, neglect higher order terms and it gives you a quick answer.
     
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