$\lim_{n \to \infty} \int_{0}^{n}(1-\frac{3x}{n})^ne^{\frac{x}{2}}dx$=? $$\lim_{n \to \infty} \int_{0}^{n}\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}}dx$$
I thought about using the theorem of monotonic convergence and had $$f_n{(x)}=\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}} \lambda_{[0,n]}(x)$$ it is rising. but I don't know where it is valid to write: 
$$\lim_{n \to \infty} \int_{0}^{n}\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}}dx= \int_{0}^{\infty}\lim_{n \to \infty}\left(1-\frac{3x}{n}\right)^ne^{\frac{x}{2}}dx=\int_{0}^{\infty}e^{\frac{x}{2}-3x}dx$$
 A: The integral diverges.  To see this, we can write
$$\int_0^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=\int_0^{n/3} \left(1-\frac{3x}n\right)^ne^{x/2}\,dx+\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx \tag 1$$
We will present two parts.  In Part $1$, we will show that the first integral on the right-hand side of $(1)$ converges.  In Part $2$, we will show that the second integral diverges.

PART $1$:
The first integral on the right-hand side of $(3)$ can easily be shown to converge.  In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to establish that the exponential function satisfies the inequality 
$$\left(1+\frac{x}{n}\right)^n\le e^x$$
for $x>-n$.  Therefore, we assert that $\left(1-\frac{3x}n\right)^n\le e^{-3x}$ for $x<n/3$.  We can then write
$$\begin{align}
\int_0^{n/3}\left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\int_0^{\infty}\left(1-\frac{3x}n\right)^n\,\lambda_{[0,n]}(x)\,e^{x/2}\,dx\\\\
&\le \int_0^\infty e^{-5x/2}\,dx\\\\
&=\frac25
\end{align}$$
The dominated convergence theorem guarantees that 
$$\begin{align}
\lim_{n\to \infty}\int_0^{n/3}\left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\int_0^\infty \lim_{n\to \infty}\left(1-\frac{3x}n\right)^n\,\lambda_{[0,n]}(x)\,e^{x/2}\,dx\\\\
&=\int_0^\infty e^{-3x}\,(1)\,e^{x/2}\,dx\\\\
&=\frac25
\end{align}$$

PART $2$:
We will now analyze the second integral on the right-hand side of $(1)$.  We enforce the substitution $x\to nx/3$ to reveal
$$\begin{align}
\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\frac{n}{3} \int_1^{3}\left(1-x\right)^ne^{nx/6}\,dx\\\\
&=(-1)^n\frac{n}{3}e^{n/6}\int_{0}^{2}x^ne^{nx/6}\,dx
\end{align}$$
Since $\int_0^2 x^ne^{nx/6}\,dx\ge \frac{2^{n+1}-1}{n+1}$, then we have
$$\lim_{n\to \infty}\int_{n/3}^n \left(1-\frac{3x}n\right)^ne^{x/2}\,dx=
\begin{cases}
\infty&, n\,\,\text{even}\\\\
-\infty&, n\,\,\text{odd}
\end{cases}$$
Therefore, the limit of interest does not exist and we cannot interchange the order of the limit and the integral.
A: Let $f_n(x)=\left(1-\frac{3x}{n}\right)^ne^{x/2}$.
Then $$\lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)=\lim_{n\to\infty }f_n(x)\cdot \underbrace{\lim_{n\to\infty }\lambda_{[0,n]}(x)}_{=\lambda_{[0,\infty [}(x)}=\lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x).$$
Therefore $$\int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x)\mathrm d x=\int_0^\infty \lim_{n\to\infty }f_n(x)\mathrm d x.$$
