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Babynini

  • one year ago

Should I use squeeze theorem? how?

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  1. Babynini
    • one year ago
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    Prove that lim (as x approaches 0 from the positive side) sq{x}[1+sin^2(2pi/)]

  2. Babynini
    • one year ago
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    @Empty :)

  3. Babynini
    • one year ago
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    @zepdrix

  4. thomas5267
    • one year ago
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    Is this the limit you want? I type the previous limit wrong. \[ \lim_{x\to0+}\sqrt{x}\sin^2(2\pi) \]

  5. Babynini
    • one year ago
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    Correct, except sin^2(2pi/x) at the end there

  6. thomas5267
    • one year ago
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    This one? \[ \lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right) \]

  7. Babynini
    • one year ago
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    Yep!

  8. Babynini
    • one year ago
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    ooh after the end there is an = 0

  9. Babynini
    • one year ago
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    Sorry sorry. So all of that that you just wrote out = 0

  10. Empty
    • one year ago
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    Yeah any time you see a sine function, you can usually squeeze it away since its range is always bounded. \[0 \le \sin^2(f(x)) \le 1\] In this case \[f(x)=\frac{2 \pi}{x}\] but it could literally be any real valued function, doesn't matter. Now we can take our inequality: \[-\sqrt{x}0 \le \sqrt{x}\sin^2(f(x)) \le \sqrt{x}\] And I think it might be easy to see how to continue with the squeeze theorem from here! The main thing to take away is that the bounded nature of sine or cosine really ends up at most contributing some constant to any limit, so you can kinda throw it away very easily in your mind once you get used to the idea that it won't contribute to your specific limit.

  11. Babynini
    • one year ago
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    on the very first equation you wrote out, did you mean to place sin between -1 and 1? instead of 0 and 1.

  12. Empty
    • one year ago
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    Nah cause I squared it :P

  13. Babynini
    • one year ago
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    Agh, wait a moment. Let me write out the original one because I think thomas forgot to put the 1+ in front of the sin in there o.o or maybe I forgot too

  14. Empty
    • one year ago
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    You have the right idea I just took your inequality that you're thinking of a step further cause \[a^2 \ge 0\]

  15. Babynini
    • one year ago
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    \[\lim_{x \rightarrow 0^+}\sqrt{x}[1+\sin^2(2\pi/x)]=0\]

  16. thomas5267
    • one year ago
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    Actually the limit is: \[ \lim_{x\to 0+}\sqrt{x}\left(1+\sin\left(\frac{2\pi}{x}\right)\right)={\lim_{x\to 0+}\sqrt{x}}+\lim_{x\to 0+}\sqrt{x}\sin\left(\frac{2\pi}{x}\right) \] The results still apply.

  17. Babynini
    • one year ago
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    ..except sin^2

  18. thomas5267
    • one year ago
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    Apparently I had a brain fart thinking even though \(0\leq\sin^2(x)\leq1\) and \(\lim_{x\to0+}0\sqrt{x}=\lim_{x\to0+}\sqrt{x}=0\) still doesn't mean that \(\lim_{x\to0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)=0\).

  19. Babynini
    • one year ago
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    But you are dropping the 1+ in front of sine!

  20. Babynini
    • one year ago
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    @Empty does the thing you did still work even with the 1+? where does it go? I can pretty easily do the theorem, only that 1 is messing me up.

  21. thomas5267
    • one year ago
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    Maximum of sin(x) is 1. For some value of x, sin(x)=0 From the maximum of sin(x) we can deduce that the maximum of sin^2(x)=1. As for some value of x sin(x)=0 so sin^2(x)=0. Since the minimum value of x^2 is 0, the minimum of sin^2(x)=(sin(x))^2=0.

  22. thomas5267
    • one year ago
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    \[\lim_{x\to 0+}\sqrt{x}\left(1+\sin^2\left(\frac{2\pi}{x}\right)\right)={\lim_{x\to 0+}\sqrt{x}}+\lim_{x\to 0+}\sqrt{x}\sin^2\left(\frac{2\pi}{x}\right)\] Both limit exist so you can expand the brackets

  23. thomas5267
    • one year ago
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    I had quite a lot of brain fart in this question.

  24. zepdrix
    • one year ago
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    \[\large\rm 0\lt \sin^2\left(\frac{2\pi}{x}\right)\le1\]\[\large\rm 1\lt 1+\sin^2\left(\frac{2\pi}{x}\right)\le2\]\[\large\rm \sqrt{x}\lt \sqrt{x}\left[1+\sin^2\left(\frac{2\pi}{x}\right)\right]\le2\sqrt{x}\]Why is the +1 messing you up nini? :o You still get zeroes on both ends of this inequality, ya?

  25. thomas5267
    • one year ago
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    *Both limits Plural singular brain fart.

  26. Babynini
    • one year ago
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    ooh it all just moves up by 1?

  27. Babynini
    • one year ago
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    hahah @thomas5267 you and I both! I'm not sure why this problem is being so troublesome xD

  28. Empty
    • one year ago
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    Maybe it'll make sense to look at the range of \(\sin x\): \[-1 \le \sin x \le 1\] The range is just \([-1,1]\) But when we square it, the negative part of the range goes away: \[0 \le \sin^2 x \le 1\] since \([0,1]\) is the new range.

  29. Babynini
    • one year ago
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    And then that range [0,1] becomes [1,2] because of the 1+ in the equation in front of sine?

  30. Babynini
    • one year ago
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    (Yes, thank you, that made it make so much more sense xD)

  31. Babynini
    • one year ago
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  32. Babynini
    • one year ago
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    is that..correct?

  33. Empty
    • one year ago
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    Yeah looks perfect to me!

  34. Empty
    • one year ago
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    Actually one tiny thing, \[0 \le \sin^2x\] since they can be equal!

  35. zepdrix
    • one year ago
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    I think my inequality was wrong. For some reason I was thinking the sine function should be restricted to (0,1] because of the x in the denominator. But clearly you can put in x=2 to get get zero :P Shouldn't be strict inequality.

  36. zepdrix
    • one year ago
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    or to get* zero, as opposed to get getting zero -_- dat typo

  37. Babynini
    • one year ago
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    Ah okay, so i just need to add the "=" to each of the "<" on the left side!

  38. Babynini
    • one year ago
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    hm, @zepdrix ?

  39. zepdrix
    • one year ago
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    ya :)

  40. Babynini
    • one year ago
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    yaay thank you guys!

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