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Squeeze theorem.

  1. Jan 6, 2008 #1
    1. The problem statement, all variables and given/known data

    evaluate the limit.

    lim x -> infinity sin2^x / x^4

    2. Relevant equations

    squeeze theorem.

    3. The attempt at a solution

    I believe the squeeze theorem could be relevant here.
    also, i know that sin 1 is 0. so couldn't this problem be something like 0 < sinx/ x < 0

    thus also making the answer 0? am i on the right track, please help, thanks.
     
  2. jcsd
  3. Jan 6, 2008 #2

    olgranpappy

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    it's not clear what you mean by sin2^x
     
  4. Jan 6, 2008 #3

    Dick

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    Not this again! This time I'll guess (sin(x))^2. Am I right?
     
  5. Jan 6, 2008 #4

    olgranpappy

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    :rofl:

    I'm going to guess sin(x^2)... let's see who's right... Of course, the answer is the same either way.
     
  6. Jan 6, 2008 #5

    olgranpappy

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    ... unfortunately he could also mean ( sin(2) )^x

    which would not have the same limit....
     
  7. Jan 6, 2008 #6

    Dick

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    Let's not forget sin(2^x).
     
  8. Jan 6, 2008 #7

    olgranpappy

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    ... so we can say that there is a 75 percent chance that the answer is zero, and a 25 percent chance it's infinity.
     
  9. Jan 6, 2008 #8
  10. Jan 6, 2008 #9

    olgranpappy

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    yes, you are on the right track.
     
  11. Jan 6, 2008 #10

    Dick

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    I win! That's sin(x)^2. So squeeze it. -1<=sin(x)<=1. What about sin(x)^2?
     
  12. Jan 6, 2008 #11
    okay, so since i know that sin is limited to 1 and -1, then also sin(x)^2 is also limited to 1?

    so which is it then, is sin(x)^2 1 or -1? i know its no more or no less than either of the two.

    thanks so far.
     
  13. Jan 6, 2008 #12

    olgranpappy

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    it is, indeed, never greater than 1.

    no... it's always more than -1... it is always the square of a real number...
     
  14. Jan 6, 2008 #13
    so then it is fair to say, it is 1.
     
  15. Jan 6, 2008 #14

    Dick

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    It's not 1. 0<=sin^2(x)<=1. Now finish the problem.
     
  16. Jan 6, 2008 #15

    Gib Z

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    Come on rcmango, it doesn't even matter really about the inequality, you could have -243425254 <= sin^2 (x) <= 429837928562983462394 for all we care. The point is the numerator is bounded. Is the denominator? What does this mean?
     
  17. Jan 7, 2008 #16
    Okay, so we know that sin^2 (x) is bounded between 0 and 1,

    the denominator, x^4 is not bounded. it just keeps getting bigger, so number could be like 1, 1/16, 1/81 and so on ... right? thus getting closer to 0. So sin^2 (x)/x^4 = 0.

    I've put alot of effort thinking about this, i'm not very good at it, thanks for the help.
     
  18. Jan 8, 2008 #17

    olgranpappy

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    you must mean to say
    [tex]
    lim_{x\to\infty}\sin^2(x)/x^4=0
    [/tex]
    , right?

    It would also probably be useful to graph the function and you will see right away that as 'x' gets large
    sin^2(x)/x^4 get very close to zero quite quickly.
     
  19. Jan 8, 2008 #18

    Gib Z

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    rcmango: Yes that is a reasonable manner to think in, but it would be better to have done it more like this;

    Since we have the strict inequality [itex] 0 \leq \sin^2 x \leq 1[/itex], we also have the inequality (provided x is not equal to zero) - [tex]0 \leq \frac{\sin^2 x}{x^4} \leq \frac{1}{x^4}[/tex]. Now taking limits as we let the value of x be arbitrarily large, we arrive at:

    [tex]0 \leq \lim_{x\to \infty} \frac{\sin^2 x}{x^4} \leq \lim_{x\to \infty} \frac{1}{x^4}[/tex]. Since the limit on the right is also equal to 0, by the squeeze theorem (which your thread name implies was originally to be used), the limit [tex]\lim_{x\to \infty} \frac{\sin^2 x}{x^4}[/tex] is equal to 0.
     
    Last edited: Jan 8, 2008
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