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moongazerBest ResponseYou've already chosen the best response.0
\[\lim_{x \rightarrow 0} \frac{ \sin (\sin x) }{ x }\] How do you find this?
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
$$sin(sin(x))=x+O(x^3)$$
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
divide by x, let x tend to zero
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
sin(sin(x))=x+O(x^3)
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
Because if $$\sin(x)=x+O(x^3)$$
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
$$sin(\sin(x))=x+O(x^3)$$
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
where did you get sin(sin(x))=x+O(x^3) ?
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
I substituted sin(x) for x
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
appearing in the first sin(x)
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
or you could look at it, and see its clearly one, sense $$sin(x)=x+O(x^3)$$
 9 months ago

TuringTestBest ResponseYou've already chosen the best response.2
@Jack17 I don;t think this is Taylor series level math yet, just precal
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
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 :)
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
So no differentiation? :(
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
its just a extra problem given by our teacher for challenge I think
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
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
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
so, there is no simple way for solving this? like using trig identities or something?
 9 months ago

terenzreignzBest ResponseYou've already chosen the best response.0
Trig identities generally don't provide for nested trig functions, like, this one, sine of the sine... At least, none that I know of...
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
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.
 9 months ago

SidKBest ResponseYou've already chosen the best response.1
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.
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.
 9 months ago

TuringTestBest ResponseYou've already chosen the best response.2
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\[\lim_{x\to0}\frac{\sin x}x\]
 9 months ago

cinarBest ResponseYou've already chosen the best response.2
can we say this sinx/x<\[ \cos x \le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
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
 9 months ago

TuringTestBest ResponseYou've already chosen the best response.2
@cinar 's inequalities are fine, I think
 9 months ago

cinarBest ResponseYou've already chosen the best response.2
I believe this is OK but I am not sure about cosx side \[\le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
nvm, I am to lazy to work this stuff out
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
I tried doing what @SidK did but I am stuck finding the lim(x > 0) (sin x)x
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
rly moongazer, there are like 50 answers here
 9 months ago

TuringTestBest ResponseYou've already chosen the best response.2
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\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
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
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
though I think by that time it would be over kill
 9 months ago

TuringTestBest ResponseYou've already chosen the best response.2
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.
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
how did you find lim(x > 0) (sin x)/x ? sorry if it is a dumb question :)
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
you could make some geometric argument, or bound it by other trigonometric functions, apply lhopitals rule, there are many ways
 9 months ago

TuringTestBest ResponseYou've already chosen the best response.2
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
 9 months ago

cinarBest ResponseYou've already chosen the best response.2
use squeeze rule lim x >0\[\cos x \le \frac{ sinx}{x } \le1\]=1
 9 months ago

cinarBest ResponseYou've already chosen the best response.2
nice tutorial https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/prooflimsinxx
 9 months ago

Jack17Best ResponseYou've already chosen the best response.0
Ye I think he does some geometric argument in that one lol
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
thanks @cinar I think that's the best solution for my level :)
 9 months ago

moongazerBest ResponseYou've already chosen the best response.0
and thank you everybody for giving different ideas on how to solve this :)
 9 months ago
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