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 (Sun):

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

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- May 9th 2008, 09:07 PMChris L T521An 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 (Sun):

$\displaystyle \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}\,dx$. - May 9th 2008, 10:43 PMmr fantastic
*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. - May 9th 2008, 10:44 PMwingless
$\displaystyle \lim_{b\to \infty}\int_{0}^{b}\frac{\sqrt{1-e^{-x}}}{b}~dx$

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

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:

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

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

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

$\displaystyle \lim_{b\to \infty}\sqrt{1-e^{-b}} = 1$ - May 9th 2008, 11:00 PMCaptainBlack
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 - May 9th 2008, 11:01 PMmr fantastic
- May 9th 2008, 11:07 PMCaptainBlack
- May 9th 2008, 11:12 PMmr fantastic
- May 11th 2008, 01:35 PMKrizalid
Let's make this more generally:

If $\displaystyle \lim_{x\to\infty}f(x)=\lambda\implies\lim_{x\to\in fty}\frac1x\int_0^x f(y)\,dy=\lambda.$ - May 11th 2008, 02:29 PMThePerfectHacker
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 $\displaystyle \lim_{x\to \infty} \frac{1}{x} \int_0^x f(\xi) d\xi = L$." - May 11th 2008, 02:46 PMThePerfectHacker
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 $\displaystyle \lim_{x\to \infty} \frac{1}{x}\int_0^x \sqrt{1-e^{-t}}dt = 1$.