## moongazer 3 years ago limit of [sin(sin x)]/x as x approaches to 0

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1. moongazer

$\lim_{x \rightarrow 0} \frac{ \sin (\sin x) }{ x }$ How do you find this?

2. anonymous

$$\sin(x)=x+O(x^3)$$

3. anonymous

$$sin(sin(x))=x+O(x^3)$$

4. anonymous

divide by x, let x tend to zero

5. moongazer

why?

6. anonymous

why what

7. moongazer

sin(sin(x))=x+O(x^3)

8. anonymous

Because if $$\sin(x)=x+O(x^3)$$

9. anonymous

$$sin(\sin(x))=x+O(x^3)$$

10. anonymous

so the limit is 1

11. moongazer

where did you get sin(sin(x))=x+O(x^3) ?

12. anonymous

I substituted sin(x) for x

13. anonymous

appearing in the first sin(x)

14. anonymous

use L'Hospital rule

15. anonymous

or you could look at it, and see its clearly one, sense $$sin(x)=x+O(x^3)$$

16. TuringTest

@Jack17 I don;t think this is Taylor series level math yet, just precal

17. anonymous

[cos(sin x)cos x]/1

18. moongazer

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

So no differentiation? :(

20. moongazer

its just a extra problem given by our teacher for challenge I think

21. moongazer

@terenzreignz yes

22. terenzreignz

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

so, there is no simple way for solving this? like using trig identities or something?

24. terenzreignz

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. anonymous

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. anonymous

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. anonymous

In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.

28. TuringTest

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. anonymous

can we say this sinx/x<$\cos x \le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1$

30. anonymous

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. anonymous

or the rhs actually.

32. TuringTest

@cinar 's inequalities are fine, I think

33. TuringTest

oh maybe not...

34. anonymous

I believe this is OK but I am not sure about cosx side $\le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1$

35. anonymous

nvm, I am to lazy to work this stuff out

36. moongazer

I tried doing what @SidK did but I am stuck finding the lim(x --> 0) (sin x)x

37. anonymous

rly moongazer, there are like 50 answers here

38. TuringTest

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. anonymous

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. anonymous

though I think by that time it would be over kill

41. TuringTest

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

how did you find lim(x --> 0) (sin x)/x ? sorry if it is a dumb question :)

43. anonymous

you could make some geometric argument, or bound it by other trigonometric functions, apply lhopitals rule, there are many ways

44. TuringTest

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. anonymous

use squeeze rule lim x ->0$\cos x \le \frac{ sinx}{x } \le1$=1

46. anonymous
47. anonymous

Ye I think he does some geometric argument in that one lol

48. moongazer

thanks @cinar I think that's the best solution for my level :)

49. moongazer

and thank you everybody for giving different ideas on how to solve this :)