# Sandwich theorem

Is it true that sin(x) <= x <= tan(x) , for x close to zero??

Can this be proven using the unit circle?

Rather then using L'hopitals rule to solve lim(x->0) of Sin(3x)/x (answer = 3)

Related Precalculus Mathematics Homework Help News on Phys.org
$$sinx\leq x$$ for $$x\in[0,\pi]$$ or

$$|sinx|\leq |x|$$ for every x. Also

$$|tanx| \geq |x|$$ for $$x \neq \frac{\pi}{2}(2k+1)$$

the proof is quite straightforward in both cases using the unit circle, although there are other methods without reference to the unite circle at all. remember that this is true only when x is measured in radian.

Last edited:
$$sinx\leq x$$ for $$x\in[0,\pi]$$
I just want to jump in here with my massive ignorance and ask something (I haven't gotten to trig yet).

Isn't sin(n) always in the range [0,1]? That is certainly how the sin function of all the programming languages I've used behaves.

If so, sin(n) is always < n in (1,->). Why stop at pi?

k

well, if we want to go all the way through we have to put it under the absolute values, like i did a line below. Because take

$$x=-\frac{\pi}{6}$$ than clearly the following is not valid

$$sin({-\frac{\pi}{6})=-\frac{1}{2}\leq -\frac{\pi}{6}$$

YOu see the point.

So basically when sin falls in the 3rd and 4rth quadrant than we are in trouble.

I just want to jump in here with my massive ignorance and ask something (I haven't gotten to trig yet).

Isn't sin(n) always in the range [0,1]? That is certainly how the sin function of all the programming languages I've used behaves.

If so, sin(n) is always < n in (1,->). Why stop at pi?

k
and one more thing

$$|sin(x)|\leq 1$$ for all x.

Oh, ok. I thought [0,pi] denoted only the values between 0 and pi inclusive, which would all be positive (except 0 I guess).

Ugh, if the notation I've been using is being thrown out the window for a new one at some point, I'll be mad. I hate when they teach you something that's "wrong" as an "just for now" sort of thing.

k

Oh, ok. I thought [0,pi] denoted only the values between 0 and pi inclusive, which would all be positive (except 0 I guess).

Ugh, if the notation I've been using is being thrown out the window for a new one at some point, I'll be mad. I hate when they teach you something that's "wrong" as an "just for now" sort of thing.

k
yeah [0,pi] includes only values between zero and pi. But i just arbitrarly took this interval, since i did not write sinx in abs values, so i wanted to make sure i am talking for only positive values.

So is the following correct??

sin(3x) <= 3x <= tan(3x)

sin(3x) <= 3x <= sin(3x)/cos(3x)

1 <= 3x/sin(3x) <= 1/cos(x)

1 => sin(3x)/3x => cos(3x)

3 => sin(3x)/x => 3cos(3x)

lim(x->0) 3 => lim(x->0) sin(3x)/x => lim(x->0) 3cos(3x)

3 => lim(x->0) sin(3x)/x => 3

.: lim(x->0) sin(3x)/x = 3 (by sandwich theorem)

Any help would be greatly appreciated.

Dick
Homework Helper
It looks correct to me, if you are careful to qualify 0<=x<pi/2. Gotta admit, tho, I think l'Hopital is a bit more direct.

Yeah, like Dick mentioned, you have to be carefull on the interval you are working on.

I thought when you used the sandwich theorem is was known that you are only talking about points close to the limit you are evaluating? So in this case only points close to 0 and yes L'Hopitals rule is much easier but i just wanted to see if it could be done a different way.

I thought when you used the sandwich theorem is was known that you are only talking about points close to the limit you are evaluating? So in this case only points close to 0 and yes L'Hopitals rule is much easier but i just wanted to see if it could be done a different way.