# Calculate $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx$ .

It is required to calculate $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx$. The following is my attempt.

$\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ for each $n\in\mathbb{N}$. Let $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ for each $n\in\mathbb{N}$ and $x\in[0,\infty)$.Then $g_n(x)\leq e^{-x}$ for each $x\in[0,\infty)$ and $n\in\mathbb{N}$. $\int_{[0,\infty)}e^{-x}dx$ exists. Moreover $g_n(x)$ converges to $g(x)=e^{-x}$ for each $x$. Then by Dominated Convergence theorem $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}e^{-x}dx$.

Could someone please tell me if my solution is alright? Thanks.

• That is absolutely fine. – Lord Shark the Unknown May 18 '17 at 3:45
• You're reasoning is solid. – Mark Viola May 18 '17 at 4:17
• Thank you very much Shark and Mark :) – Janitha357 May 18 '17 at 7:52

The reasoning is correct. There is only one small typo in the definition of $g_n$: it should be $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}$ instead of $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$.