$\lim_{n \to \infty} \int_{0}^{n}(1-\frac{3x}{n})^ne^{\frac{x}{2}}dx$=?

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.

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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

$$\begin{align} \lim_{n\to \infty}\int_0^{n/3}\left(1-\frac{3x}n\right)^ne^{x/2}\,dx&=\int_0^\infty \lim_{n\to \infty}\left(1-\frac{3x}n\right)^n\,\lambda_{[0,n]}(x)\,e^{x/2}\,dx\\\\ &=\int_0^\infty e^{-3x}\,(1)\,e^{x/2}\,dx\\\\ &=\frac25 \end{align}$$

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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.

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 $$\int \lim_{n\to\infty }f_n(x)\lambda_{[0,n]}(x)\mathrm d x=\int \lambda_{[0,\infty [}(x)\lim_{n\to\infty }f_n(x)\mathrm d x=\int_0^\infty \lim_{n\to\infty }f_n(x)\mathrm d x.$$