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Does the Limit exist?

  1. May 25, 2013 #1
    1. lim as x->∞of 3x3+cosx/sinx-x3


    2. My Attempt below:


    3. lim as x->∞ of 3x3+lim as x->∞ of cosx/lim as x->∞ of sinx-lim as x->∞ of x3

    Breaking down problem even further:

    lim as x->∞ of 3x3 = ∞
    lim as x->∞ of cosx= 1. This is because by squeeze theorem, -1<cosx<1, therefore lim reaches 1
    lim as x->∞ of sinx= 1. This is because by squeeze theorem, -1<sinx<1, therefore lim reaches 1
    lim as x->∞ of x3 =∞

    ∴ = +1/1-∞ = -∞

    Is this correct?
     
  2. jcsd
  3. May 25, 2013 #2
    I don't think you understand the squeeze theorem. Do you mean ##\frac{3x^3+\cos{x}}{\sin{x}-x^3}## or ##3x^3 + \frac{\cos{x}}{\sin{x}-x^3}##?
     
  4. May 25, 2013 #3
    ##\frac{3x^3+\cos{x}}{\sin{x}-x^3}##
     
  5. May 25, 2013 #4
    ##\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}##
    the limit does exist
     
  6. May 25, 2013 #5
    And does the limit = -∞
     
  7. May 25, 2013 #6
    no, sorry, it's

    ##\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}=-3##

    try with the squeeze theorem
     
  8. May 25, 2013 #7
    I am not getting the concept of squeeze theorem clearly. I thought the following:

    lim as x->∞ of cosx= 1. This is because by squeeze theorem, -1<cosx<1, therefore lim reaches 1
    lim as x->∞ of sinx= 1. This is because by squeeze theorem, -1<sinx<1, therefore lim reaches 1

    What am I doing wrong?
     
  9. May 25, 2013 #8
    You aren't understanding what the squeeze rule says.

    Let ##g(x)\leq f(x)\leq h(x)##. The squeeze rule says that, if ##\displaystyle \lim_{x\rightarrow\alpha}g(x)=\lim_{x\rightarrow\alpha}h(x)=L##, then ##\displaystyle \lim_{x\rightarrow\alpha}f(x)=L##.

    For example, the squeeze theorem doesn't apply for ##\displaystyle \lim_{x\rightarrow\infty}\sin{x}## because ##\displaystyle \lim_{x\rightarrow\infty}[-1]\neq\lim_{x\rightarrow\infty}[1]##.

    Edit:
    Think of it this way.

    $$\lim_{x\rightarrow\infty}\frac{3x^3+\cos{x}}{\sin{x}-x^3}=\lim_{x\rightarrow\infty}\frac{3x^3(1+\frac{\cos{x}}{3x^3})}{-x^3(-\frac{\sin{x}}{x^3}+1)}$$
     
    Last edited: May 25, 2013
  10. May 25, 2013 #9
    ok then. When applying the squeeze theorem,

    how are you concluding to this?


    ##\lim_{x\to\infty} \frac{3x^3+\cos(x)}{\sin(x)-x^3}=-3##
     
  11. May 25, 2013 #10
    so ...

    $$\lim_{x\rightarrow\infty}\frac{3x^3+\cos{x}}{\sin{x}-x^3}=\lim_{x\rightarrow\infty}\frac{3x^3(1+\frac{\cos{x}}{3x^3})}{-x^3(-\frac{\sin{x}}{x^3}+1)}$$ = -3

    I GET IT! @Mandelbroth. your new edit was helpful! VERY HELPFUL

    Cheers,
     
  12. May 26, 2013 #11

    HallsofIvy

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    I'm glad you got it. Now, let me point out that [itex]\lim_{x\to \infty} sin(x)[/itex] and [itex]\lim_{x\to \infty} cos(x)[/itex] are NOT 0. They continue to alternate between -1 and 1 and do not "approach" any number. You use the "squeeze theorem" to argue that because [itex]-1\le sin(x)\le 1[/itex] and [itex]-1\le cos(x)\le 1[/itex], [itex]-\frac{1}{x}\le \frac{sin(x)}{x}\le \frac{1}{x}[/itex] and [itex]-\frac{1}{x}\le \frac{sin(x)}{x}\le \frac{1}{x}[/itex] so that [itex]\lim_{x\to\infty} \frac{sin(x)}{x}= \lim_{x\to\infty} \frac{cos(x)}{x}= 0[/itex].
     
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