Computing limit using Dominated Convergence Theorem

Question

Compute $\lim\limits_{k\to\infty} \int_0^k(1 + {\frac{x}{k}})^k e^{-2x}dx$

I tried to find a dominating fuction and found that because of

$\left(1+\frac xn\right)^n\leqslant e^x$ we can conclude $|f_n(x)|\leqslant e^{-x}$ but whats bothering me most of the time ist the $k$ in the integral as I don't know how to handle this.

I also tried to use the Binomial theorem but couldn't produce much.

Answer 1

Your integrand is actually $\left(1+\frac xn\right)^ne^{-2x}1_{0\leq x\leq k}$. But you have that $1_{0\leq x\leq k}\leq 1$. Now you can conclude using your argument

Answer 2

Hint. One may write $$ \int_0^k \left(1 + {\frac{x}{k}}\right)^k e^{-2x}dx=\int_0^\infty \mathbb{1}_{[0,k]}(x)\cdot\left(1 + {\frac{x}{k}}\right)^k e^{-2x}dx. $$