Answer
converges to ${\dfrac{1}{e^5}}$.
Work Step by Step
As we know that a sequence converges when $\lim\limits_{n \to \infty}a_n$ exists.
Consider $a_n=(\dfrac{n-5}{n})^n$
Re-write the given sequence as:$a_n=(1-\dfrac{5}{n})^n$
Apply limits to both sides.
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}(1-\dfrac{5}{n})^n$
Put $y=\lim\limits_{n \to \infty}(1-\dfrac{5}{n})^n$
Use logarithmic rule: $\ln a^n=n \ln a$
$\ln y=\lim\limits_{n \to \infty}n \ln (1-\dfrac{5}{n})$
$\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}\frac{0}{0}$
Since, we can see that the limit has the form of $\frac{0}{0}$, so take the help of L-Hospital's rule.
$\ln y=\lim\limits_{n \to \infty} \dfrac{\ln (1-5/n)}{1/n}$
$\ln y=\lim\limits_{n \to \infty} \dfrac{-5/ (1-5/n)}{1}$
$\ln y=-5 \implies y={\dfrac{1}{e^5}}$
Therefore, the sequence converges to ${\dfrac{1}{e^5}}$.
