Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

divide by x, let x tend to zero

Answer 5

why?

Answer 6

why what

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

so the limit is 1

Answer 11

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

Answer 12

I substituted sin(x) for x

Answer 13

appearing in the first sin(x)

Answer 14

use L'Hospital rule

Answer 15

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

Answer 16

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

Answer 17

[cos(sin x)cos x]/1

Answer 18

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 :)

Answer 19

So no differentiation? :(

Answer 20

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

Answer 21

@terenzreignz yes

Answer 22

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

Answer 23

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

Answer 24

Trig identities generally don't provide for nested trig functions, like, this one, sine of the sine...

At least, none that I know of...

Answer 25

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.

Answer 26

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.

Answer 27

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

Answer 28

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\[\lim_{x\to0}\frac{\sin x}x\]

Answer 29

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

Answer 30

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

Answer 31

or the rhs actually.

Answer 32

@cinar 's inequalities are fine, I think

Answer 33

oh maybe not...

Answer 34

I believe this is OK but I am not sure about cosx side \[\le\frac{ \sin(sinx)}{x } \le \frac{ sinx}{x } \le1\]

Answer 35

nvm, I am to lazy to work this stuff out

Answer 36

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

Answer 37

rly moongazer, there are like 50 answers here

Answer 38

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\[\lim_{x\to0}\frac{\sin(\sin x)}{x}\to\lim_{x\to0}\frac{\sin x}{x}\to\lim_{x\to0}\frac{x}{x}=1\]

Answer 39

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

Answer 40

though I think by that time it would be over kill

Answer 41

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.

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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