Answer
$\lim\limits_{n \to \infty} b_n=0$
Work Step by Step
Definition of a sequence defined by a function: Let us consider a function $f(x)$ whose limit $\lim\limits_{x \to \infty} f(x)$ exists then the sequence $b_n=f(n)$ will converge to the same limit.
That is, $\lim\limits_{n \to \infty} b_n=\lim\limits_{x \to \infty} f(x)$
Here, in this problem we have $b_n=ne^{-n}$ and $f(x)=\dfrac{x}{e^x}$
Next, $\lim\limits_{n \to \infty} b_n=\lim\limits_{x \to \infty} \dfrac{x}{e^x}$
In order to compute the limit, we will apply L-Hospital's Rule.
$\lim\limits_{n \to \infty} b_n=-\lim\limits_{x \to \infty} \dfrac{1}{e^x}$
Since, $n \gt 0$ for all the values of $n$ and $\lim\limits_{n \to \infty} e^n=\infty$ and $\lim\limits_{n \to \infty} e^{-n}=0$
Thus. $\lim\limits_{n \to \infty} b_n=0$