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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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  2. zepdrix
    • one year ago
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    You don't have to do this step, but it might make things easier. I would recommend making a substitution. \(\large\rm \dfrac{1}{t}=u\). Then notice that as \(\large\rm t\to0^-\), we have \(\large\rm u\to -\infty\) I can explain that further if it's confusing ^. So our limit becomes:\[\large\rm \lim_{t\to0^-}\sin^2\left(\frac{1}{t}\right)3^{1/t}=\lim_{u\to-\infty}\sin^2(u)3^u\]And then Squeeze Theorem is a little easier from this point.

  3. anonymous
    • one year ago
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    I don't think my teacher wants me to make the substitution but for me it makes it easier so far!

  4. zepdrix
    • one year ago
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    Either way is fine, let's stick with the original then. With Squeeze Theorem, you want to start with only your "troublesome piece" and try to get some bounds on it. In this problem, the sine is what's causing trouble for us. It fluctuates back and forth forever as t gets closer to 0. So let's put some hard boundaries on sine and continue from there.

  5. anonymous
    • one year ago
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    ok

  6. zepdrix
    • one year ago
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    Well we know that sine is stuck between -1 and 1. But let's just right to sine squared if we can. That gets rid of all of the negative values,\[\large\rm 0\le \sin^2\left(\frac{1}{t}\right)\le 1\]

  7. zepdrix
    • one year ago
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    Multiply all sides by 3^(1/t),\[\large\rm 0\cdot3^{1/t}\le \sin^2\left(\frac{1}{t}\right)3^{1/t}\le 1\cdot3^{1/t}\]\[\large\rm 0\le \sin^2\left(\frac{1}{t}\right)3^{1/t}\le 3^{1/t}\]

  8. zepdrix
    • one year ago
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    From here, if you can show that the left and right most sides of this inequality converge to 0, then the stuff in the middle has to converge to 0 by the Squeeze Theorem. The left-most side is pretty straight-forward, ya?\[\large\rm \lim_{t\to0^-}0=0\]

  9. zepdrix
    • one year ago
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    For the right-most side, set up the limit and do something to show it's zero. Maybe that u-sub if it makes it easier.

  10. anonymous
    • one year ago
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    how do i show that the right side is 0 though without usub?

  11. anonymous
    • one year ago
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    left side makes sense though haha

  12. anonymous
    • one year ago
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    this whole sandwich theorem is all new to me so im still trying to figure everything out haha

  13. zepdrix
    • one year ago
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    Sandwiches are delicious :o so don't let it scare you. We're just trying to show that as t approaches 0 from the left, our inequality is approaching: \(\large\rm 0\le stuff\le 0\) The stuff is sandwiched between 0 and 0, so it has to be 0.

  14. zepdrix
    • one year ago
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    Without u-sub? Ummm

  15. zepdrix
    • one year ago
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    Probably with some simple words to justify it... like umm

  16. anonymous
    • one year ago
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    its just that we haven't been taught it so i don't want him to get upset or anything

  17. anonymous
    • one year ago
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    maybe as t gets closer to 0, the right side approaches 0? so since there both =0 the original limit is 0?

  18. zepdrix
    • one year ago
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    You're just restating that the right limit is 0 not "showing" it yet lol but that's ok, it's hard to say how thorough your teacher wants you to be. Hopefully you can make some sense out of why it should be zero though.\[\Large\rm \lim_{t\to0^-}3^{1/t}\]Let's plug a number really close to zero in, that's below zero, just to get an idea of what is going on,\[\Huge\rm 3^{\frac{1}{\color{orangered}{t}}}\approx3^{\frac{1}{\color{orangered}{-\frac{1}{99999}}}}\]We're dividing by a fraction in the exponent, so we can flip it,\[\large\rm =3^{-99999}\]Rule of exponents lets us write it like this,\[\large\rm \frac{1}{3^{99999}}\]Which is a really really small number, almost 0.

  19. zepdrix
    • one year ago
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    I could have chosen a decimal I suppose like t=-0.0000001, but fractions are easier to flip.

  20. zepdrix
    • one year ago
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    But ya maybe just use some words and teach will be ok with it :) Since \(\large\rm 0\to0\) and \(\large\rm 3^{1/t}\to0\) as \(\large\rm t\to0^-\), by the Squeeze Theorem we can conclude that \(\large\rm \sin^2\left(\frac{1}{t}\right)3^{1/t}\to0\) as well.

  21. anonymous
    • one year ago
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    ok i see what your doing here

  22. anonymous
    • one year ago
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    thanks!

  23. anonymous
    • one year ago
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    reagarding the bounds, why is it 0 and 1?

  24. zepdrix
    • one year ago
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    Are you ok with the normal boundaries of the sine function? Like if I wrote this\[\large\rm -1\le \sin(x)\le 1\]does it make sense?

  25. anonymous
    • one year ago
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    ya

  26. anonymous
    • one year ago
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    sin^2 doesn't though...

  27. zepdrix
    • one year ago
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    Well, when you square something, it eliminates any negative values, ya?

  28. zepdrix
    • one year ago
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    |dw:1444368409918:dw|

  29. zepdrix
    • one year ago
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    |dw:1444368473226:dw|

  30. zepdrix
    • one year ago
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    It can't give us anything negative, so we lose half of that range from the sine function.

  31. anonymous
    • one year ago
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    ok makes sense! thanks

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