Interesting Limit (20)

It is easy to show that $e^x \ge 1+x$ for all real $x$ , and hence that $0 \le 1-x \le e^{-x}$ for all $0 \le x \le 1$ , Thus, if we define $f_n(x) \; = \; \left\{ \begin{array}{lll} \big(1 - \tfrac{x}{n}\big)^n e^{\frac12x} & \hspace{1cm} & 0 \le x \le n \\ 0 & & x > n \end{array} \right.$ then $f_n \in L^1(0,\infty)$ for all positive integers $n$ , and $0 \le f_n(x) \le g(x) \hspace{2cm} \lim_{n \to \infty}f_n(x) \; =\; g(x)$ for all $x \ge 0$ , where $g(x) = e^{-\frac12x}$ . Hence, by the Dominated Convergence Theorem, $\lim_{n \to \infty} \int_0^\infty f_n(x)\,dx \; = \; \int_0^\infty g(x)\,dx \; = \; \boxed{2}$