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Proof of limit by definition

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    find the limit as x -> 0 of (sin^2(x))/(x^2)

    2. Relevant equations

    limit as x -> xo (fx) = L iff for every epsilon (>0) there exists a delta (>0) st if
    | x - xo | < delta then |f(x) - L | < epsilon



    3. The attempt at a solution

    Let epsilon be positive. I believe the limit equals one, so I will proceed there.
    then

    | (sin^2(x))/x^2 - 1| < epsion if | x | < delta

    But | f(x) - 1| <= |1/x^2 - 1| . And this is where I get stuck. If I pick delta to be small, and |x| < delta,
    | (f(x)) - 1 | becomes very large, and thus is not being bound by any epsilon.
     
  2. jcsd
  3. Oct 26, 2008 #2

    Dick

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    |f(x)-1| does NOT become very large as x->0. Did you experiment with some numbers, like x=0.01,x=0.0001, etc? Do you know anything about the limit of sin(x)/x as x->0?
     
  4. Oct 26, 2008 #3
    You are right, I meant to say that

    |1/x^2 - 1| becomes large as x -> 0, which confused me, because

    | f(x) - 1| <= |1/x^2 - 1|, which i get because | sin^2(x) | <= 1, so then | sin^2(x)/x^2 - 1 | <= |1/x^2 - 1 |


    The limit of sin(x)/x equals 1 when x approaches 0. Are you suggesting that I use the fact that
    sin(x)/x has limit 1 at x = 0 and the theorem that if lim f exists and equals L1 at some point xo and lim g exists at the same xo and equals L2, then lim f*g = L1*L2, which then I could use to lim sin^2(x)/x^2 =
    lim (sin(x)/x)*(sin(x)/x) = 1 * 1 = 1. Yes, but then I need to prove that lim sin(x)/x equals 1. How would I do this by epsilon-delta definition of limits ( I know how to with squeeze theorem)? I would basically be stuck in the same mess I have above.
     
  5. Oct 26, 2008 #4

    Dick

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    |sin(x)^2/x^2-1|<=|1/x^2-1| is just not a very good estimate of |f(x)-1|. How do you prove sin(x)/x=1 using the squeeze theorem? You should be able to express that logic in epsilon-delta form. You might also notice |f(x)-1|=|sin(x)/x-1|*|sin(x)/x+1|.
     
  6. Oct 26, 2008 #5
    | cos(x) | <= | (sin x)/x | <= | 1 |
    Is how I would use the squeeze theorem to solve this one. But then I have no idea how to get from there to a delta, or reduction of | (sin x)/x | to | x | from which I can pull a delta. I have been searching the web for a few hours and have found 0 proofs that (sin x)/x has limit equal to 1. There was a place that had it listed as an exercise of finding deltas, but gave no explanation of how to find one. So I remain stuck.

    One other idea I had is
    | sin(x)/x | <= | sin(x) | <= | x |

    Then you get stuck by the triangle inequality trying to bring back (sin x)/x - 1, the closest I got is:
    | sin(x)/x - 1 + 1 | <= | sin(x)/x - 1 + 1 | <= | x - 1 + 1 |
     
  7. Oct 26, 2008 #6

    Dick

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    I think I'm better at searching the web than you are. I found this:

    Multiply 1-cos(x) by (1+cos(x))/(1+cos(x)) and get (1-cos(x)^2)/(1+cos(x))=sin(x)^2/(1+cos(x))<=sin(x)^2 (for small x). You also have sin(x)<=x. So put it all together and get 1-cos(x)<=x^2. So:

    1-x^2<=sin(x)/x<=1. Does that look like a form you can use?
     
  8. Oct 26, 2008 #7

    Oh, I never said I was good at searching the web, only that I had done it for a while...

    Anyways, I like where you went with this. Basically:

    |sin x| < |x| . This implies that |(sin x)/x| < |x/x| = 1.

    On the other hand, if x is in the open interval (0,pi/2) then | x | < | tan x |, so we get that

    | x/sin(x) | < | (tan x) / (sin x) | = | 1/(cos x) |. Then you reciprocate the rationals over the inequality to get that:

    | cos x | < | (sin x) / x|. Then you put the two of these together to get:

    | cox x | < | (sin x) / x| < 1. (1)

    Now for the next bit, we will need again that (sin anynumber) < anynumber, and the famous identity sin^2(x) + cos^2(x) = 1, and then we can use your trick:
    (1- cos x) * (1 + cos x)/(1 + cos x) = (1 - (cos x)^2)/(1+ cos x) = ((sin x)^2)/(1+ cos x) <= (sin x)^2
    But because (sin x) < x, we get that (1 - cos x) < (x)^2. Then:
    - cos x < x^2 -1 (2) , and further
    cos x > 1 - x^2. Now because we have limited x to the open interval (0,pi/2), and from the inequalities (1) and (2)

    1 - x^2 < (sin x)/x < 1 , and now the big one
    - x^2 < (sin x)/x - 1 <0 (3)and zero is obviously less than epsilon (which is chosen positive). However, this is not complete, the above will not hold on all intervals, particularly the above holds when x is in (0,pi/2) in other words | x | < pi/2. Notice that,
    (sin x) / x - 1 < x^2
    And so if delta equals sqrt(epsilon) we get that if | x | < delta
    then (sin x) / x - 1 < x^2 < epsilon
    We need to pick delta to be the min of pi/2 and sqrt(epsilon).
     
  9. Oct 26, 2008 #8

    Dick

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    Yes, I think that does it. You shouldn't have any troubles extending that to get the delta for sin(x)^2/x^2.
     
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