limit of [sin(sin x)]/x as x approaches to 0

- moongazer

limit of [sin(sin x)]/x as x approaches to 0

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- moongazer

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

- anonymous

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

- anonymous

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

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## More answers

- anonymous

divide by x, let x tend to zero

- moongazer

why?

- anonymous

why what

- moongazer

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

- anonymous

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

- anonymous

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

- anonymous

so the limit is 1

- moongazer

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

- anonymous

I substituted sin(x) for x

- anonymous

appearing in the first sin(x)

- anonymous

use L'Hospital rule

- anonymous

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

- TuringTest

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

- anonymous

[cos(sin x)cos x]/1

- 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

So no differentiation? :(

- moongazer

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

- moongazer

@terenzreignz yes

- 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

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

- 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

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

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

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

- 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

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

- 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

or the rhs actually.

- TuringTest

@cinar 's inequalities are fine, I think

- TuringTest

oh maybe not...

- 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

nvm, I am to lazy to work this stuff out

- moongazer

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

- anonymous

rly moongazer, there are like 50 answers here

- 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

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

though I think by that time it would be over kill

- 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

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

- anonymous

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

- 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

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

- anonymous

nice tutorial
https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x

- anonymous

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

- moongazer

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

- moongazer

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

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