## moongazer Group Title limit of [sin(sin x)]/x as x approaches to 0 one year ago one year ago

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1. moongazer Group Title

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

2. Jack17 Group Title

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

3. Jack17 Group Title

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

4. Jack17 Group Title

divide by x, let x tend to zero

5. moongazer Group Title

why?

6. Jack17 Group Title

why what

7. moongazer Group Title

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

8. Jack17 Group Title

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

9. Jack17 Group Title

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

10. Jack17 Group Title

so the limit is 1

11. moongazer Group Title

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

12. Jack17 Group Title

I substituted sin(x) for x

13. Jack17 Group Title

appearing in the first sin(x)

14. cinar Group Title

use L'Hospital rule

15. Jack17 Group Title

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

16. TuringTest Group Title

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

17. cinar Group Title

[cos(sin x)cos x]/1

18. moongazer Group Title

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 Group Title

So no differentiation? :(

20. moongazer Group Title

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

21. moongazer Group Title

@terenzreignz yes

22. terenzreignz Group Title

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 Group Title

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

24. terenzreignz Group Title

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 Group Title

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 Group Title

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 Group Title

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

28. TuringTest Group Title

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 Group Title

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

30. Jack17 Group Title

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 Group Title

or the rhs actually.

32. TuringTest Group Title

@cinar 's inequalities are fine, I think

33. TuringTest Group Title

oh maybe not...

34. cinar Group Title

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 Group Title

nvm, I am to lazy to work this stuff out

36. moongazer Group Title

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

37. Jack17 Group Title

rly moongazer, there are like 50 answers here

38. TuringTest Group Title

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 Group Title

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 Group Title

though I think by that time it would be over kill

41. TuringTest Group Title

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 Group Title

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

43. Jack17 Group Title

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

44. TuringTest Group Title

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 Group Title

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

46. cinar Group Title
47. Jack17 Group Title

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

48. moongazer Group Title

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

49. moongazer Group Title

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