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Thread: 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 :


    $\displaystyle \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx$.
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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 :


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

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

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

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

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

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

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

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

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

    Use L'hopital:

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

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

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

    $\displaystyle \lim_{b\to \infty}\sqrt{1-e^{-b}} = 1$
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  4. #4
    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 :


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

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

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

    So

    $\displaystyle
    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
    $

    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:

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

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

    So

    $\displaystyle
    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
    $

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

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

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

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

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

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

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

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