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moongazer

  • one year ago

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

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

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

  3. Jack17
    • one year ago
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    $$sin(sin(x))=x+O(x^3)$$

  4. Jack17
    • one year ago
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    divide by x, let x tend to zero

  5. moongazer
    • one year ago
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    why?

  6. Jack17
    • one year ago
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    why what

  7. moongazer
    • one year ago
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    sin(sin(x))=x+O(x^3)

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

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

  10. Jack17
    • one year ago
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    so the limit is 1

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

  12. Jack17
    • one year ago
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    I substituted sin(x) for x

  13. Jack17
    • one year ago
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    appearing in the first sin(x)

  14. cinar
    • one year ago
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    use L'Hospital rule

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

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

  17. cinar
    • one year ago
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    [cos(sin x)cos x]/1

  18. moongazer
    • one year ago
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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 :)

  19. terenzreignz
    • one year ago
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    So no differentiation? :(

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

  21. moongazer
    • one year ago
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    @terenzreignz yes

  22. terenzreignz
    • one year ago
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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

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

  24. terenzreignz
    • one year ago
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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...

  25. Jack17
    • one year ago
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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.

  26. SidK
    • one year ago
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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.

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

  28. TuringTest
    • one year ago
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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\[\lim_{x\to0}\frac{\sin x}x\]

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

  30. Jack17
    • one year ago
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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

  31. Jack17
    • one year ago
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    or the rhs actually.

  32. TuringTest
    • one year ago
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    @cinar 's inequalities are fine, I think

  33. TuringTest
    • one year ago
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    oh maybe not...

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

  35. Jack17
    • one year ago
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    nvm, I am to lazy to work this stuff out

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

  37. Jack17
    • one year ago
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    rly moongazer, there are like 50 answers here

  38. TuringTest
    • one year ago
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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\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]

  39. Jack17
    • one year ago
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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

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

  41. TuringTest
    • one year ago
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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.

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

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

  44. TuringTest
    • one year ago
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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

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

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

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

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

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

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