moongazer
  • moongazer
limit of [sin(sin x)]/x as x approaches to 0
Discrete Math
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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moongazer
  • moongazer
\[\lim_{x \rightarrow 0} \frac{ \sin (\sin x) }{ x }\] How do you find this?
anonymous
  • anonymous
$$\sin(x)=x+O(x^3)$$
anonymous
  • anonymous
$$sin(sin(x))=x+O(x^3)$$

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anonymous
  • anonymous
divide by x, let x tend to zero
moongazer
  • moongazer
why?
anonymous
  • anonymous
why what
moongazer
  • moongazer
sin(sin(x))=x+O(x^3)
anonymous
  • anonymous
Because if $$\sin(x)=x+O(x^3)$$
anonymous
  • anonymous
$$sin(\sin(x))=x+O(x^3)$$
anonymous
  • anonymous
so the limit is 1
moongazer
  • moongazer
where did you get sin(sin(x))=x+O(x^3) ?
anonymous
  • anonymous
I substituted sin(x) for x
anonymous
  • anonymous
appearing in the first sin(x)
anonymous
  • anonymous
use L'Hospital rule
anonymous
  • anonymous
or you could look at it, and see its clearly one, sense $$sin(x)=x+O(x^3)$$
TuringTest
  • TuringTest
@Jack17 I don;t think this is Taylor series level math yet, just precal
anonymous
  • anonymous
[cos(sin x)cos x]/1
moongazer
  • 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 :)
terenzreignz
  • terenzreignz
So no differentiation? :(
moongazer
  • moongazer
its just a extra problem given by our teacher for challenge I think
moongazer
  • moongazer
@terenzreignz yes
terenzreignz
  • 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
moongazer
  • moongazer
so, there is no simple way for solving this? like using trig identities or something?
terenzreignz
  • 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...
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.
TuringTest
  • 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\[\lim_{x\to0}\frac{\sin x}x\]
anonymous
  • anonymous
can we say this sinx/x<\[ \cos x \le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]
anonymous
  • 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
anonymous
  • anonymous
or the rhs actually.
TuringTest
  • TuringTest
@cinar 's inequalities are fine, I think
TuringTest
  • TuringTest
oh maybe not...
anonymous
  • anonymous
I believe this is OK but I am not sure about cosx side \[\le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]
anonymous
  • anonymous
nvm, I am to lazy to work this stuff out
moongazer
  • moongazer
I tried doing what @SidK did but I am stuck finding the lim(x --> 0) (sin x)x
anonymous
  • anonymous
rly moongazer, there are like 50 answers here
TuringTest
  • 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\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]
anonymous
  • 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
anonymous
  • anonymous
though I think by that time it would be over kill
TuringTest
  • 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.
moongazer
  • moongazer
how did you find lim(x --> 0) (sin x)/x ? sorry if it is a dumb question :)
anonymous
  • anonymous
you could make some geometric argument, or bound it by other trigonometric functions, apply lhopitals rule, there are many ways
TuringTest
  • 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
anonymous
  • anonymous
use squeeze rule lim x ->0\[\cos x \le \frac{ sinx}{x } \le1\]=1
anonymous
  • anonymous
nice tutorial https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x
anonymous
  • anonymous
Ye I think he does some geometric argument in that one lol
moongazer
  • moongazer
thanks @cinar I think that's the best solution for my level :)
moongazer
  • moongazer
and thank you everybody for giving different ideas on how to solve this :)

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