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

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

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

divide by x, let x tend to zero

why?

why what

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

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

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

so the limit is 1

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

I substituted sin(x) for x

appearing in the first sin(x)

use L'Hospital rule

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

[cos(sin x)cos x]/1

So no differentiation? :(

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

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

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

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

or the rhs actually.

@cinar 's inequalities are fine, I think

oh maybe not...

nvm, I am to lazy to work this stuff out

rly moongazer, there are like 50 answers here

though I think by that time it would be over kill

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

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

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

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