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Math Help - An Interesting Limit...

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Talking An Interesting Limit...

    I already evaluated the limit, and got the answer, but I want to see if you guys can come up with an easier way to solve it :


    \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx.
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  2. #2
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    Quote Originally Posted by Chris L T521 View Post
    I already evaluated the limit, and got the answer, but I want to see if you guys can come up with an easier way to solve it :


    \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx.
    *Yawn* I'll bite.

    \frac{\int_{0}^{b}\sqrt{1-e^{-x}}\,dx}{b} is an indeterminant form \left( \frac{\infty}{\infty}\right) so I'll just be lazy and apply l'Hospital:

    \lim_{b\to \infty} \frac{\sqrt{1-e^{-b}}}{1} = 1.

    There are many other ways - I'll save them for the pleasure of others.
    Last edited by mr fantastic; May 9th 2008 at 10:58 PM. Reason: Added last line
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  3. #3
    Super Member wingless's Avatar
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    \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}~dx

    \lim_{b\to \infty}\frac{\int_{0}^{b}\sqrt{1-e^{-x}}~dx}{b}

    Now we know that \int_{0}^{b}\sqrt{1-e^{-x}}~dx converges because it's always greater than zero and the limit of \sqrt{1-e^{-x}} is 1. So the limit is in indeterminate form \frac{\infty}{\infty}. We can use L'hopital later.

    Define F(x) = \int \sqrt{1-e^{-x}}~dx.

    \int_{0}^{b}\sqrt{1-e^{-x}}~dx = F(b) - F(0)

    Now it's \lim_{b\to \infty}\frac{F(b) - F(0)}{b}.

    Use L'hopital:

    \lim_{b\to \infty}\frac{\frac{d}{db}F(b) - \frac{d}{db}\not{F(0)}}{\frac{d}{db}~b}

    \lim_{b\to \infty}\frac{\frac{d}{db}F(b)}{1}

    \lim_{b\to \infty}\frac{d}{db}\int \sqrt{1-e^{-b}}~db

    \lim_{b\to \infty}\sqrt{1-e^{-b}} = 1
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    Grand Panjandrum
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    Quote Originally Posted by Chris L T521 View Post
    I already evaluated the limit, and got the answer, but I want to see if you guys can come up with an easier way to solve it :


    \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx.
    For once L'hopitals's rule is appropriate:

    L=\lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx=\lim_{b \to \infty}\frac{\int_0^{b}\sqrt{1-e^{-x}}~dx}{b}.

    (Note the integral in the expression on the right goes to infinity as b does as the integrand goes to 1 for large x)

    So

     <br />
L=\lim_{b \to \infty} \frac{(d/db)\left(\int_0^{b}\sqrt{1-e^{-x}}~dx \right)}{(d/db)(b)}=\lim_{b \to \infty} \frac{\sqrt{1-e^{-b}}}{1}=1<br />

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    For once L'hopitals's rule is appropriate:

    L=\lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx=\lim_{b \to \infty}\frac{\int_0^{b}\sqrt{1-e^{-x}}}{b}.

    (Note the integral in the expression on the right goes to infinity as the integrand goes to 1 for large x)

    So

     <br />
L=\lim_{b \to \infty} \frac{(d/db)\left(\int_0^{b}\sqrt{1-e^{-x}}\right)}{(d/db)(b)}=\lim_{b \to \infty} \frac{\sqrt{1-e^{-b}}}{1}=1<br />

    RonL
    The l'Hospital's busy tonight


    By the way, CaptainB ..... "For once L'hopitals's rule is appropriate" .... solid gold!
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by mr fantastic View Post
    The l'Hospital's busy tonight


    By the way, CaptainB ..... "For once L'hopitals's rule is appropriate" .... solid gold!

    Careless of me, I hadn't noticed the other replies

    I take that back, just looked at the cached version of the question, no replies when I started typing, so just slow typing on my or other's part

    RonL
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  7. #7
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    Quote Originally Posted by CaptainBlack View Post
    Careless of me, I hadn't noticed the other replies

    RonL
    Well, if you hadn't replied, I wouldn't be wiping away tears of laughter right now.
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  8. #8
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    Let's make this more generally:

    If \lim_{x\to\infty}f(x)=\lambda\implies\lim_{x\to\in  fty}\frac1x\int_0^x f(y)\,dy=\lambda.
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  9. #9
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    Quote Originally Posted by Krizalid View Post
    Let's make this more generally:

    If \lim_{x\to\infty}f(x)=\lambda\implies\lim_{x\to\in  fty}\frac1x\int_0^x f(y)\,dy=\lambda.
    It is not as general as you wrote it. You need to satisfy some conditions.

    "If f:[0,\infty)\mapsto \mathbb{R} is a continous function and \lim_{x\to \infty} f(x) exists and is L, and if \int_0^{\infty} f(\xi) d\xi = \infty then \lim_{x\to \infty} \frac{1}{x} \int_0^x f(\xi) d\xi = L."
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  10. #10
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    Here is a solution without L'Hopital's rule.

    Note that, 1 - e^{-t} \leq \sqrt{1-e^{-t}} \leq 1 for all t\geq 0.

    Thus, \int_0^x 1 - e^{-t} dt \leq \int_0^x \sqrt{1-e^{-t}}dt \leq \int_0^x 1dx.

    This gives us,  x + (e^{-x} - 1) \leq  \int_0^x \sqrt{1-e^{-t}}dt \leq x.

    Dividing by x gives, 1 + \frac{e^{-x} - 1}{x} \leq \frac{1}{x}\int_0^x \sqrt{1-e^{-t}}dt \leq 1.

    Since \lim_{x\to \infty} \frac{e^{-x} - 1}{x} = 0 it follows by the Squeeze theorem that \lim_{x\to \infty} \frac{1}{x}\int_0^x \sqrt{1-e^{-t}}dt = 1.
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