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Homework Help: Sandwich theorem

  1. Apr 7, 2008 #1
    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)
  2. jcsd
  3. Apr 8, 2008 #2
    [tex] sinx\leq x[/tex] for [tex]x\in[0,\pi][/tex] or

    [tex]|sinx|\leq |x|[/tex] for every x. Also

    [tex] |tanx| \geq |x| [/tex] for [tex] x \neq \frac{\pi}{2}(2k+1)[/tex]

    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: Apr 8, 2008
  4. Apr 8, 2008 #3
    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?

  5. Apr 8, 2008 #4
    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

    [tex] x=-\frac{\pi}{6}[/tex] than clearly the following is not valid

    [tex] sin({-\frac{\pi}{6})=-\frac{1}{2}\leq -\frac{\pi}{6}[/tex]

    YOu see the point.

    So basically when sin falls in the 3rd and 4rth quadrant than we are in trouble.
  6. Apr 8, 2008 #5
    and one more thing

    [tex]|sin(x)|\leq 1[/tex] for all x.
  7. Apr 8, 2008 #6
    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.

  8. Apr 8, 2008 #7
    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.
  9. Apr 8, 2008 #8
    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.
  10. Apr 8, 2008 #9


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    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.
  11. Apr 8, 2008 #10
    Yeah, like Dick mentioned, you have to be carefull on the interval you are working on.
  12. Apr 8, 2008 #11
    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.

    Thanks for your help!
  13. Apr 8, 2008 #12
    Yeah, when we use the sandwich theorem, we pretty much look only for points close enought to some point c, but from both sides, that is from the left and from the right also. But since you have x close to zero, it means that x can be negative also, which would make the following statement untrue

    sinx<x , say x=-b, where b>0 so

    sin(-b)<-b=>-sin(b)<-b=>sin(b)>b, which is not true, here this is a small contradiction, or maybe a warning that we have either to make sure x is only positive, or to put sin and x in abs values.
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