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moongazer

limit of [sin(sin x)]/x as x approaches to 0

  • 9 months ago
  • 9 months ago

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  1. moongazer
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    \[\lim_{x \rightarrow 0} \frac{ \sin (\sin x) }{ x }\] How do you find this?

    • 9 months ago
  2. Jack17
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    $$\sin(x)=x+O(x^3)$$

    • 9 months ago
  3. Jack17
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    $$sin(sin(x))=x+O(x^3)$$

    • 9 months ago
  4. Jack17
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    divide by x, let x tend to zero

    • 9 months ago
  5. moongazer
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    why?

    • 9 months ago
  6. Jack17
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    why what

    • 9 months ago
  7. moongazer
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    sin(sin(x))=x+O(x^3)

    • 9 months ago
  8. Jack17
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    Because if $$\sin(x)=x+O(x^3)$$

    • 9 months ago
  9. Jack17
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    $$sin(\sin(x))=x+O(x^3)$$

    • 9 months ago
  10. Jack17
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    so the limit is 1

    • 9 months ago
  11. moongazer
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    where did you get sin(sin(x))=x+O(x^3) ?

    • 9 months ago
  12. Jack17
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    I substituted sin(x) for x

    • 9 months ago
  13. Jack17
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    appearing in the first sin(x)

    • 9 months ago
  14. cinar
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    use L'Hospital rule

    • 9 months ago
  15. Jack17
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    or you could look at it, and see its clearly one, sense $$sin(x)=x+O(x^3)$$

    • 9 months ago
  16. TuringTest
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    @Jack17 I don;t think this is Taylor series level math yet, just precal

    • 9 months ago
  17. cinar
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    [cos(sin x)cos x]/1

    • 9 months ago
  18. moongazer
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    we still didn't discuss L'Hospital rule we have only discuses up to squeeze theorem yes this is still precalculus so I think I am missing something :)

    • 9 months ago
  19. terenzreignz
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    So no differentiation? :(

    • 9 months ago
  20. moongazer
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    its just a extra problem given by our teacher for challenge I think

    • 9 months ago
  21. moongazer
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    @terenzreignz yes

    • 9 months ago
  22. terenzreignz
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    Pity, I was hoping to do this: \[\Large \lim_{x\rightarrow 0}\frac{\sin[\sin(x)]}{x}=\lim_{x\rightarrow 0}\frac{\sin[\sin(x)]-\sin[\sin(0)]}{x} \] \[\Large \left.\frac{d}{dx}\sin[\sin(x)]\right|_{x=0}\] Oh well... rethinking :D

    • 9 months ago
  23. moongazer
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    so, there is no simple way for solving this? like using trig identities or something?

    • 9 months ago
  24. terenzreignz
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    Trig identities generally don't provide for nested trig functions, like, this one, sine of the sine... At least, none that I know of...

    • 9 months ago
  25. Jack17
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    That would depend on your definition of simple, but anyone who knows the behaviour of sin(x) near zero could look at that limit, and solve it on the spot.

    • 9 months ago
  26. SidK
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    Note that lim(x-->0) sin t / t = 1. So, lim(x-->0) sin(sin x) / x = lim(x-->0) [sin(sin x) / sin x] * [sin x / x]. As x --> 0, sin x -->0. Hence, lim(x-->0) [sin(sin x) / sin x] = 1. So, the final answer is 1 * 1 = 1.

    • 9 months ago
  27. Jack17
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    In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.

    • 9 months ago
  28. TuringTest
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    according to the argument above As x --> 0, sin x -->0. Hence, lim(x-->0) [sin(sin x) / sin x] = 0/0 I would make a verbal argument about how \[\sin x\approx x\] for small x, so\[\sin \sin x\approx \sin x\] for small enough x. Thise two ideas lead to just\[\lim_{x\to0}\frac{\sin x}x\]

    • 9 months ago
  29. cinar
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    can we say this sinx/x<\[ \cos x \le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]

    • 9 months ago
  30. Jack17
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    Yes, I think we have solved this problem in about 30 ways, also I am not sure if that inequality is true on the lhs

    • 9 months ago
  31. Jack17
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    or the rhs actually.

    • 9 months ago
  32. TuringTest
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    @cinar 's inequalities are fine, I think

    • 9 months ago
  33. TuringTest
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    oh maybe not...

    • 9 months ago
  34. cinar
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    I believe this is OK but I am not sure about cosx side \[\le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]

    • 9 months ago
  35. Jack17
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    nvm, I am to lazy to work this stuff out

    • 9 months ago
  36. moongazer
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    I tried doing what @SidK did but I am stuck finding the lim(x --> 0) (sin x)x

    • 9 months ago
  37. Jack17
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    rly moongazer, there are like 50 answers here

    • 9 months ago
  38. TuringTest
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    whatever, my solution: for small x, we have that \(\sin x\approx x\), therefore for small \(\sin x\) we have \(\sin(\sin x)\approx\sin x\), so as x gets small we get\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]

    • 9 months ago
  39. Jack17
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    you could make this more rigorous with error bounds using taylors theorem or bounding off the summands occuring in the maclaurin series of sin(x) past x

    • 9 months ago
  40. Jack17
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    though I think by that time it would be over kill

    • 9 months ago
  41. TuringTest
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    Yeah, I think we need to stay in the realm of precalc here, so I tried to stay minimal. My proof is not rigorous to be sure.

    • 9 months ago
  42. moongazer
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    how did you find lim(x --> 0) (sin x)/x ? sorry if it is a dumb question :)

    • 9 months ago
  43. Jack17
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    you could make some geometric argument, or bound it by other trigonometric functions, apply lhopitals rule, there are many ways

    • 9 months ago
  44. TuringTest
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    That can be proven with L'hospital, Taylor expansion, or geometrical argument. I am taking it as a given that they already know that limit, though I do point out again that \(\sin x\approx x\) for small enough x

    • 9 months ago
  45. cinar
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    use squeeze rule lim x ->0\[\cos x \le \frac{ sinx}{x } \le1\]=1

    • 9 months ago
  46. cinar
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    nice tutorial https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x

    • 9 months ago
  47. Jack17
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    Ye I think he does some geometric argument in that one lol

    • 9 months ago
  48. moongazer
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    thanks @cinar I think that's the best solution for my level :)

    • 9 months ago
  49. moongazer
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    and thank you everybody for giving different ideas on how to solve this :)

    • 9 months ago
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