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Sandwich theorem limit problem

  1. Dec 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that
    $$ \lim_{x\to 0} \sqrt{x^3+x^2}\; \sin\left(\frac{\pi}{x}\right) = 0 $$

    using Sandwich theorem
    2. Relevant equations
    Sandwich Theorem

    3. The attempt at a solution
    Now we know that sine function takes values between -1 and 1. ## -1 \leqslant \sin\left(\frac{\pi}{x}\right) \leqslant 1 ##. So we can multiply this by ## \sqrt{x^3+x^2} ## on both sides. But I want to show that ## \sqrt{x^3+x^2} \geqslant 0## near ## x = 0##. So I am stuck at this point. Any guidance will help.

    Thanks
     
  2. jcsd
  3. Dec 28, 2016 #2

    Delta²

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    For x>0 its easy. Now for x<0 if you take for example ##-1<x<0## to prove that ##x^3>-x^2##. Hint: try multiplying ##-1<x<0## by the proper positive term)
     
  4. Dec 28, 2016 #3
    Though I am solving this problem from Stewart's Calculus book, I am trying to be rigorous here. In the statement of the Sandwich theorem, we define an interval ##I## having ##0## as its limit point. Now if we take ##I = [-1,1]##, then ##0## is definitely the limit point of ##[-1,1]##. Now we need to show that for every x in ##I## not equal to ##0##, we have
    $$ -\sqrt{x^3 + x^2} \leqslant \sin\left(\frac{\pi}{x}\right) \leqslant \sqrt{x^3 + x^2}$$

    For this, we will first need to show that ##\sqrt{x^3 + x^2} \geqslant 0## for all ##x## in ##[-1,1]##. Am I on right track ?
     
  5. Dec 28, 2016 #4

    Delta²

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    yes you are on the right track ( you just forgot to multiply by the ##\sqrt{x^3+x^2}## term in the middle of the double inequality).
     
  6. Dec 28, 2016 #5
    Oh I missed that.. Ok I will continue working on this.... thanx
     
  7. Dec 28, 2016 #6

    Ray Vickson

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    There is nothing to prove; that is the very definition of the function ##\sqrt{...} .##
     
  8. Dec 28, 2016 #7
    Ray, we need to be sure that ## x^3+x^2 \geqslant 0##. If this is not true, then there might be problem. Thats why I am tried to come up with an interval over which, we have ## x^3+x^2 \geqslant 0##, and this interval also has ##0## as its limit point.
     
  9. Dec 28, 2016 #8

    Ray Vickson

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    For ##0 \leq |x| \leq 1## we have ##|x|^3 \leq |x|^2 = x^2##.
     
  10. Dec 28, 2016 #9
    Thanks Ray... makes sense
     
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