1. ## 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 :

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

2. Originally Posted by Chris L T521
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.

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

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

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:

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

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

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

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

4. Originally Posted by Chris L T521
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

$
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

5. Originally Posted by CaptainBlack
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

$
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!

6. Originally Posted by mr fantastic
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

7. Originally Posted by CaptainBlack
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.

8. Let's make this more generally:

If $\lim_{x\to\infty}f(x)=\lambda\implies\lim_{x\to\in fty}\frac1x\int_0^x f(y)\,dy=\lambda.$

9. Originally Posted by Krizalid
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 $\lim_{x\to \infty} \frac{1}{x} \int_0^x f(\xi) d\xi = L$."

10. 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 $\lim_{x\to \infty} \frac{1}{x}\int_0^x \sqrt{1-e^{-t}}dt = 1$.