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moongazer

  • 2 years ago

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

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

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

  3. Jack17
    • 2 years ago
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    $$sin(sin(x))=x+O(x^3)$$

  4. Jack17
    • 2 years ago
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    divide by x, let x tend to zero

  5. moongazer
    • 2 years ago
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    why?

  6. Jack17
    • 2 years ago
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    why what

  7. moongazer
    • 2 years ago
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    sin(sin(x))=x+O(x^3)

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

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

  10. Jack17
    • 2 years ago
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    so the limit is 1

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

  12. Jack17
    • 2 years ago
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    I substituted sin(x) for x

  13. Jack17
    • 2 years ago
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    appearing in the first sin(x)

  14. cinar
    • 2 years ago
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    use L'Hospital rule

  15. Jack17
    • 2 years 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
    • 2 years ago
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    @Jack17 I don;t think this is Taylor series level math yet, just precal

  17. cinar
    • 2 years ago
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    [cos(sin x)cos x]/1

  18. moongazer
    • 2 years 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
    • 2 years ago
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    So no differentiation? :(

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

  21. moongazer
    • 2 years ago
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    @terenzreignz yes

  22. terenzreignz
    • 2 years 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
    • 2 years ago
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    so, there is no simple way for solving this? like using trig identities or something?

  24. terenzreignz
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.

  28. TuringTest
    • 2 years 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
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    or the rhs actually.

  32. TuringTest
    • 2 years ago
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    @cinar 's inequalities are fine, I think

  33. TuringTest
    • 2 years ago
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    oh maybe not...

  34. cinar
    • 2 years 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
    • 2 years ago
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    nvm, I am to lazy to work this stuff out

  36. moongazer
    • 2 years 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
    • 2 years ago
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    rly moongazer, there are like 50 answers here

  38. TuringTest
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    though I think by that time it would be over kill

  41. TuringTest
    • 2 years 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
    • 2 years ago
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    how did you find lim(x --> 0) (sin x)/x ? sorry if it is a dumb question :)

  43. Jack17
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    use squeeze rule lim x ->0\[\cos x \le \frac{ sinx}{x } \le1\]=1

  46. cinar
    • 2 years 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
    • 2 years ago
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    Ye I think he does some geometric argument in that one lol

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

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

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