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Find the limit of the function

  1. Oct 18, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the limit of the function ( attached)

    2. Relevant equations


    3. The attempt at a solution

    Is the limit infinity. Found the integral as (-1/(2t))*e^-2t from 0 to 0.
    Isn't the integral of a function that is 0 to 0, 0. The lim of 1/x approaching zero infinity.
     

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  2. jcsd
  3. Oct 18, 2016 #2

    LCKurtz

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    Apparently you are just plugging in ##x=0##, and getting ##0## for the integral and ##0## for the denominator. In other words, a ##\frac 0 0## form. You can't draw any conclusion from that. But you likely have studied techniques to handle ##\frac 0 0## forms, right?
     
  4. Oct 18, 2016 #3
    Look, if you put ##x=0##, you have ##\frac{0}{0}## form [##\int_0^x e^{-t^2}dt## is zero when ##x=0##]. So in that case, you cannot conclude that the limit is infinity.
    Now, look at the figure:
    Untitled.png
    If ##x\rightarrow 0##, what will be ##\frac{\int_0^x e^{-t^2}dt}{x}=\frac{Area}{x}## ? [Use trapezoidal approximation of area under the curve, that is, if ##x## is small enough, the area under the curve can be approximated by the area of the trapezoid ##ABCD##]
     
    Last edited: Oct 18, 2016
  5. Oct 18, 2016 #4

    LCKurtz

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    The integrand is ##e^{-t^2}##, not ##e^{-2t}## so the area can't be calculated directly.
     
  6. Oct 18, 2016 #5
    Thanks for pointing out the mistake. I did not mean to analytically calculate the area, I meant to use the trapezoidal approximation for the area.
     
  7. Oct 19, 2016 #6

    SammyS

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    Hello Sam Donovan. Welcome to PF !

    It's a lot easier to follow the discussion with the following being visible.
    larson-question-png.107691.png
     
  8. Oct 19, 2016 #7

    Ray Vickson

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    If you look at the graph of the function ##e^{-t^2}## you will see that the trapezoidal area ##T(x)## underestimates the exact area ##A(x)##, so
    $$\frac{T(x)}{x} < \frac{A(x)}{x}.$$
    Thus, you can conclude that ##\lim_{t \to 0} (1/x) \int_0^x f(t) \, dt \geq \lim_{x \to 0} T(x)/x##, and so get a lower bound on your desired limit. How can you get an upper bound as well, in order to extract an exact value?
     
  9. Oct 19, 2016 #8
    $$e^{-t^2}\leq1$$
    $$\int^x_0e^{-t^2}\,dt\leq\int^x_0\,dt$$
    $$\frac{\int^x_0e^{-t^2}\,dt}{x}\leq\frac{\int^x_0\,dt}{x}$$
     
  10. Oct 19, 2016 #9

    Ray Vickson

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    Right!

    An even easier (but looser) lower bound than the trapezoidal one is obtainable from the fact that for ##t \in (0,x)## we have ##e^{-t^2}> e^{-x^2}##, so that ##\int_0^x e^{-t^2} \, dt > \int_0^t e^{-x^2} \, dt = x e^{-x^2}##. Besides, for some other, similar, functions, the trapezoidal estimate produces an upper bound, rather than a lower bound, so you are left unable to make a safe conclusion about the limit; for example because you do not have both a lower and upper bound. For example, consider ##\int_0^x e^{-\sqrt{t}} \, dt## for small ##x > 0##.
     
    Last edited: Oct 20, 2016
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