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Question about substitution in limits?

  1. Sep 15, 2011 #1
    okay so I just have a question about using substitution when solving limits.

    Say I have the function sinx/x, if i want to find the limit when x--> 0 using approximation ( i know how to prove it with the pinching theorem by the way). So if I substitute 0.0000000001 rad into the function i get something close to 1. However if i substitute that many DEGREES I am waaaaaaaayyyyy off. I know a rad is alot more than a degree but if i draw the graphs for both wouldnt i see that the limit is 1 for both the rad graf and the degree. (This isnt an actual question where i have to use substitution but i was just experimenting and am kinda wondering...).
    any explanations would be appreciated.

    thank you.
     
    Last edited: Sep 15, 2011
  2. jcsd
  3. Sep 15, 2011 #2
    Are you sure you're expressing _both_ sinx and x in degrees when you do your calculation?
     
  4. Sep 15, 2011 #3

    mathman

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    sinx/x -> 1, only when the argument of the sin is in radians. Mathematically, the argument for all trig functions (when being considered as functions) has to be in radians.
     
  5. Sep 15, 2011 #4
    What I meant is you can look at working with degrees as just rescaling by 2Pi, i.e., start with 360=2Pi , and rescale any angle . If x->0 , then x':=2Pix/360 is a rescaling, and
    sinx'/x' also goes to 1 .
     
  6. Sep 15, 2011 #5
    yeah thats what i figured but is there a reason or explanation for it?
     
  7. Sep 15, 2011 #6

    Char. Limit

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    sin(xr) and sin(xd) (where xr and xd represent "radian x" and "degree x") are different functions, that's all. Namely, sin(xd) = sin(pi/180 xr). And so reasonably, the two functions would approach zero at different "speeds".
     
  8. Sep 15, 2011 #7
    Re: Different Speeds (My 'quote' is not working)

    Char Limit: the rescaling here makes no difference at the end, e.g., try

    L'Hopital:

    Sin(kx)/kx --> kCos(kx)/k= Cos(kx) , goes to 1 as x->0 , since k<oo ; just use, e.g

    Taylor's Thm.
     
  9. Sep 15, 2011 #8

    Char. Limit

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    Incorrect. It makes no difference if you scale x similarly, but just using the "degree sine" function without scaling x WILL make it change. We know this:

    [tex]\lim_{x \to 0} \frac{sin_r(x)}{x} = 0[/tex]

    where sinr(x) represents the "radian sine" function, which is the usual sine function. For reference, sind(x), representing the "degree sine" function, is equal to sinr(pi/180 x). So therefore:

    [tex]\lim_{x \to 0} \frac{sin_d(x)}{x} = \frac{sin_r\left(\frac{\pi}{180} x\right)}{x} = \frac{\pi}{180}[/tex]

    You see why, I hope.
     
  10. Sep 15, 2011 #9
    I only referred to a scaling, not a change into degrees; still, sinx is defined in R, i.e.,

    for all reals.

    And what's with the 'I Hope' ?
     
  11. Sep 15, 2011 #10
    And, BTW, I clearly stated that the rescaling should be done both with the argument

    and with x, and my argument reflects that ; I think you misunderstood/misinterpreted

    my answer.
     
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