Evaluate :

$a)$ $I=\lim \limits_{n\to\infty}n(1-\int_0^{n}(1-\frac{x}{n})^ndx)$

$b)$ $\lim \limits_{n\to\infty}(2-\int_0^{n}(1-\frac{x}{n})^ndx)^n$

About The first limit :

My idea is :

$I=\lim \limits_{n\to\infty}\frac{1-\int_0^{n}(1-\frac{x}{n})^ndx}{\frac{1}{n}}$

In form


I think use stolze Cesaro theorem

But the second limit I don't have any ideas to approach it!

  • $\begingroup$ I can give some hint, just for the ease of computations. for the second limit rewrite $ (2-\int_{0}^{n}(1-x/n)^n dx)^{n}=e^{n \ln (2-\int_{0}^{n}(1-x/n)dx}$, and go back to the first case.)) $\endgroup$ – kolobokish Jul 2 '19 at 16:36
  • $\begingroup$ You can write the part within the integral as e^-x. First one on simplification you get n/e^n=0 and second one you get e^(n/e^n)=1 $\endgroup$ – user600016 Jul 2 '19 at 16:48
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    $\begingroup$ What prevents you to evaluate the integral directly? $\endgroup$ – metamorphy Jul 2 '19 at 17:03

Before we bring the big guns, the change of variables $u=1-\frac{x}{n}$ gives

$$ \int^n_0(1-\frac{x}{n})^n\,dx=n\int^1_0u^ndu=\frac{n}{n+1} $$ That will simplify things; for instance,


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  • $\begingroup$ Thanks. , sir what about if we can't find integral ? $\endgroup$ – Tomas Houbaze Jul 2 '19 at 18:33
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    $\begingroup$ Then you switch from a butter knife to a Ginsu knife, and if that does not work then we try more powerful tools. There are no recipes, one only use the tools at our disposal. It was a lucky strike to notice that the problem was not as difficult as it originally looked. I just learn for instance about the Stolze--Cesàro theorem through you. $\endgroup$ – Oliver Diaz Jul 4 '19 at 17:04

For part 1, we can just directly evaluate the integral. Don't be tempted by the familiar expression to write $(1-\frac{x}{n})^n = e^{-x}$. It's quite straightforward, using the substitution $u = 1 - \frac{x}{n} \implies du = -\frac{dx}{n}$. So the integral becomes

$$I =\int_0^n \big(1- \frac{x}{n}\big)^n dx = -n \int_1^0 u^n du = \frac{n}{n+1}$$ Then we have $\lim_{n \to \infty} n (1 - \int_0^n \big(1- \frac{x}{n}\big)^n dx) = \lim_{n \to \infty} \frac{n}{n+1} = \boxed{1}$.

For part 2, we use what we calculated in part one and we have $2 - I = \frac{n+2}{n+1}$. Then the limit becomes $L = \displaystyle \lim_{n \to \infty} \bigg( \frac{n+2}{n+1} \bigg)^n \implies \ln L = \lim_{n \to \infty} \frac{(\ln(n+2) - \ln(n+1))}{\frac{1}{n}}$. By L'Hopitale, we have $\ln L = \displaystyle \lim_{n \to \infty} \frac{\frac{1}{n+2} - \frac{1}{n+1}}{-1/n^2} = \lim_{n \to \infty} \frac{\frac{-1}{(n+1)(n+2)}}{-1/n^2} = 1$. So $L = e^{\ln L} = \boxed{e}$.

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  • $\begingroup$ Thanks. , sir what about if we can't find integral ? $\endgroup$ – Tomas Houbaze Jul 2 '19 at 18:33
  • $\begingroup$ This problem is heavily contingent upon what the integral evaluates to. You would have to specify what the integral is, and we'd probably have to find an ad hoc method for evaluating this. As you guessed, using the Stolz-Cesaro Theorem could be very helpful. $\endgroup$ – paulinho Jul 2 '19 at 20:36

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