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moongazer
limit of [sin(sin x)]/x as x approaches to 0
\[\lim_{x \rightarrow 0} \frac{ \sin (\sin x) }{ x }\] How do you find this?
$$sin(sin(x))=x+O(x^3)$$
divide by x, let x tend to zero
Because if $$\sin(x)=x+O(x^3)$$
$$sin(\sin(x))=x+O(x^3)$$
where did you get sin(sin(x))=x+O(x^3) ?
I substituted sin(x) for x
appearing in the first sin(x)
or you could look at it, and see its clearly one, sense $$sin(x)=x+O(x^3)$$
@Jack17 I don;t think this is Taylor series level math yet, just precal
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 :)
So no differentiation? :(
its just a extra problem given by our teacher for challenge I think
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
so, there is no simple way for solving this? like using trig identities or something?
Trig identities generally don't provide for nested trig functions, like, this one, sine of the sine... At least, none that I know of...
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.
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.
In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.
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\[\lim_{x\to0}\frac{\sin x}x\]
can we say this sinx/x<\[ \cos x \le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]
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
@cinar 's inequalities are fine, I think
I believe this is OK but I am not sure about cosx side \[\le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]
nvm, I am to lazy to work this stuff out
I tried doing what @SidK did but I am stuck finding the lim(x --> 0) (sin x)x
rly moongazer, there are like 50 answers here
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\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]
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
though I think by that time it would be over kill
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.
how did you find lim(x --> 0) (sin x)/x ? sorry if it is a dumb question :)
you could make some geometric argument, or bound it by other trigonometric functions, apply lhopitals rule, there are many ways
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
use squeeze rule lim x ->0\[\cos x \le \frac{ sinx}{x } \le1\]=1
nice tutorial https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x
Ye I think he does some geometric argument in that one lol
thanks @cinar I think that's the best solution for my level :)
and thank you everybody for giving different ideas on how to solve this :)