Evaluate $\lim\limits_{n\to +\infty}\int\limits_{[0,n]}(1-\frac{x}{n})e^{\frac{x}{2}}dx$

Question

Problem:

Evaluate $\lim\limits_{n\to +\infty}\int\limits_{[0,n]}(1-\frac{x}{n})e^{\frac{x}{2}}dx$

I know that I should use Lebesgue's Dominated Convergence or bounded Convergence theorem for $(1-\frac{x}{n})e^{\frac{x}{2}}\chi_{[0,n]}(x)$. But I cant determine the boundedness. The measurability of the $(1-\frac{x}{n})e^{\frac{x}{2}}\chi_{[0,n]}(x)$ can be obtained from its continuity.

Answer 1

\begin{align*} \int_0^n \left(1-\frac{x}{n}\right)e^{x/2} \, dx & = 2e^{x/2}\bigg|_{x=0}^{x=n} - 2e^{x/2} \cdot\frac{x}{n}\bigg|_{x=0}^{x=n}+\frac{1}{n}\int_0^n 2e^{x/2} \, dx\\[10pt] &=2e^{n/2}-2-2e^{n/2}+ \left.\frac{4}{n}e^{x/2}\right|_{x=0}^{x=n} \\[10pt] &=-2+\frac{4}{n} e^{n/2}-\frac{4}{n}\\[10pt] &\rightarrow\infty. \end{align*}

Answer 2

Here is an approach that uses the dominated convergence theorem.

Let $f_n(x) = (1-{x \over n})e^{x \over 2}$.

Note that for some $m$ if $n \ge m$ we have $\int_0^m f_n \le \int_0^n f_n$. Dominated convergence shows that $\lim_n \int_0^m f_n = 2(e^{m \over 2} -1)$.

Hence for any $K>0$ we can pick some $m$ such that $2(e^{m \over 2} -1) > 2K$ and then select $N$ such that if $n \ge N$ we have $K \le \int_0^m f_n \le \int_0^n f_n$.

Hence $\lim_n \int_0^m f_n = \infty$.

Answer 3

The given integral is greater than

$$\int_0^{n/2} (1-x/n)e^{x/2}\, dx > \int_0^{n/2} (1/2)\cdot 1\, dx = n/4 \to \infty.$$

Although I suspect $(1-x/n)$ should have been $(1-x/n)^n$ in this problem.