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

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

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0$$sin(sin(x))=x+O(x^3)$$

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0divide by x, let x tend to zero

moongazer
 one year ago
Best ResponseYou've already chosen the best response.0sin(sin(x))=x+O(x^3)

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0Because if $$\sin(x)=x+O(x^3)$$

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0$$sin(\sin(x))=x+O(x^3)$$

moongazer
 one year ago
Best ResponseYou've already chosen the best response.0where did you get sin(sin(x))=x+O(x^3) ?

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0I substituted sin(x) for x

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0appearing in the first sin(x)

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0or you could look at it, and see its clearly one, sense $$sin(x)=x+O(x^3)$$

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

moongazer
 one year ago
Best ResponseYou've already chosen the best response.0we 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
 one year ago
Best ResponseYou've already chosen the best response.0So no differentiation? :(

moongazer
 one year ago
Best ResponseYou've already chosen the best response.0its just a extra problem given by our teacher for challenge I think

terenzreignz
 one year ago
Best ResponseYou've already chosen the best response.0Pity, 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
 one year ago
Best ResponseYou've already chosen the best response.0so, there is no simple way for solving this? like using trig identities or something?

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

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0That 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.

SidK
 one year ago
Best ResponseYou've already chosen the best response.1Note 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.

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0In general $$\lim_{x\to a} f(g(x))\ne f(\lim_{x\to a} g(x))$$.

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2according 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\[\lim_{x\to0}\frac{\sin x}x\]

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

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I think we have solved this problem in about 30 ways, also I am not sure if that inequality is true on the lhs

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2@cinar 's inequalities are fine, I think

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

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0nvm, I am to lazy to work this stuff out

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

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0rly moongazer, there are like 50 answers here

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2whatever, 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\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0you 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

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0though I think by that time it would be over kill

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2Yeah, 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
 one year ago
Best ResponseYou've already chosen the best response.0how did you find lim(x > 0) (sin x)/x ? sorry if it is a dumb question :)

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0you could make some geometric argument, or bound it by other trigonometric functions, apply lhopitals rule, there are many ways

TuringTest
 one year ago
Best ResponseYou've already chosen the best response.2That 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

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

cinar
 one year ago
Best ResponseYou've already chosen the best response.2nice tutorial https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/prooflimsinxx

Jack17
 one year ago
Best ResponseYou've already chosen the best response.0Ye I think he does some geometric argument in that one lol

moongazer
 one year ago
Best ResponseYou've already chosen the best response.0thanks @cinar I think that's the best solution for my level :)

moongazer
 one year ago
Best ResponseYou've already chosen the best response.0and thank you everybody for giving different ideas on how to solve this :)
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