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Find $L=\lim\limits_{n \to \infty} \int_0^{n a} \exp\left(-\dfrac{t}{1+\frac{b t}{n}}\right) dt$, where $a>0$, $b>0$.

I can't see what is dominating function, but I feel that I have to use Dominated convergence theorem. Any kind of help is welcome.

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up vote 1 down vote accepted

Rewrite the integral as

$$\int_0^{n a} dt \: \exp\left(-\dfrac{t}{1+\frac{b t}{n}}\right) = n a \int_0^1 du \: \exp\left(-\dfrac{n a u}{1+a b u}\right)$$

As $n \rightarrow \infty$, the contributions from the integral come mainly from the neighborhood near $u=0$. In this limit, then

$$L = \lim_{n \rightarrow \infty} n a \int_0^{\infty} du \: e^{- n a u} = 1$$

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i have a small q about your ans: i to got to the point where the integral of u is between 0 and 1/ in the next step you take L to be the same integral but u is now from o to infinity. why is that? is that a mistake? because my ans is 0... thanks! – user61202 Feb 6 '13 at 11:03
We can set the upper limit to $\infty$ because the error in doing so is exponentially small as $n \rightarrow \infty$. In fcat, you can see that in evaluating the integral with the finite limit. – Ron Gordon Feb 6 '13 at 11:12

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