Find the limit of $a_n=\left(\frac{n-4}{n}\right)^{1-n}$

Question

I was trying to solve the following task and I even got the solution but I do not understand one step in the solution.

The task was to evaluate the limit of $$a_n=\left(\frac{n-4}{n}\right)^{1-n}$$

This is how far I got:

\begin{align} a_n & =\left(\frac{n-4}{n}\right)^{1-n} \\ & = \left(1-\frac{4}{n}\right)^{1-n} \\ & = \left(1+\frac{-4}{n}\right)^{1-n} \\ & = \frac{\left(1+\frac{-4}{n}\right)}{\left(1+\frac{-4}{n}\right)^n}\\ \end{align}

This, so far is equal to the solution but the solution just simply says the following:

$$\lim_{n\to \infty}\frac{\left(1+\frac{-4}{n}\right)}{\left(1+\frac{-4}{n}\right)^n} =\frac{1}{e^{-4}}$$

I do know that $$\lim_{n\to \infty}\left(1\ +\frac{1}{n}\right)^n = e$$ But how could I get to this? $$\lim_{n\to \infty}\left(1\ +\frac{-4}{n}\right)^n = e^{-4}$$

Answer 1

If you place $\displaystyle\frac{-4}{n} = \displaystyle\frac{1}{m}$ you obtain that $n = -4m$ so when $n \to \infty$ also $m \to \infty$. Hence: $$\lim_{n \to \infty}\left(1+\frac{-4}{n}\right)^n=\lim_{m \to \infty}\left(1+\frac{1}{m}\right)^{-4m}=\lim_{m \to \infty}\left(\left(1+\frac{1}{m}\right)^{m}\right)^{-4}=e^{-4}$$

In general if you have $\lim_{n \to \infty}\left(1+\frac{a}{n}\right)^n$ you could use the substitution $\displaystyle\frac{a}{n} = \displaystyle\frac{1}{m}$ to obtain that:

$$\lim_{n \to \infty}\left(1+\frac{a}{n}\right)^n = e^a$$

Answer 2

It happens that, for each number $x$,$$\lim_n\left(1+\frac xn\right)^n=e^x.$$

Answer 3

Note simply that

$$\left(1+\frac{-4}{n}\right)^n = \left[\left(1+\frac{-4}{n}\right)^{\frac{n}{-4}}\right]^{-4} \to e^{-4}$$

Indeed

$$\lim_{x\to \pm \infty}\left(1\ +\frac{1}{x}\right)^x = e$$