How do I find $\lim_{n\to\infty}(\frac{n-1}{n})^n$

How would I find the limit for:

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

I know it approaches $\frac{1}{e}$, but I have no idea how it works. Plus, why does: $$\lim_{n\to\infty}\left(\frac{n-x}{n}\right)^n=\frac{1}{e^x}$$

Solution 1:

1. (See martini's comment). For the first question write $\frac{n-1}{n}$ as $$\begin{equation*} \frac{n-1}{n}=\frac{1}{\frac{n}{n-1}}. \end{equation*}$$ Apply limits and use the definition of $\mathrm{e}$ you have been given $$\begin{equation*} \mathrm{e}=\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n}. \end{equation*}$$ We have $$\begin{eqnarray*} \lim_{n\rightarrow \infty }\left( \frac{n-1}{n}\right) ^{n} &=&\lim_{n\rightarrow \infty }\frac{1}{\left( \frac{n}{n-1}\right) ^{n}}= \frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n}{n-1}\right) ^{n}} \\ &=&\frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n+1}}= \frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n}\cdot\lim_{n\rightarrow \infty }\frac{n+1}{n}} \\ &=&\frac{1}{\mathrm{e}\cdot 1}=\frac{1}{\mathrm{e}}. \end{eqnarray*}$$
2. Hint for the second question: write $$\begin{equation*} \left( \frac{n-x}{n}\right) ^{n}=\left( \left( 1-\frac{1}{n/x}\right) ^{n/x}\right) ^{x}, \end{equation*}$$ use the substitution $m=n/x$ and apply limits.

Solution 2:

One of the basic properties of $e$ is that $$ \lim_{ n \to \infty} \left(1+\frac{1}{n} \right)^n=e$$

You can use this here to find your answer.

Solution 3:

Here's a good strategy, first do the limit of the log of expression.

Let $y = \left(1 - \frac{x}{n} \right)^n$ (which is the same as your expression). Then, take the natural log of both sides to get $$\ln y = \ln \left[\left(1 - \frac{x}{n} \right)^n \right] = n \cdot \ln \left(1 - \frac{x}{n} \right) = \frac{\ln \left(1 - \frac{x}{n} \right)}{\frac{1}{n}}$$ Now, as $n \to \infty$, the final fraction goes to $\frac{0}{0}$, an indeterminate form. This suggests that we try l'Hopital's rule. That is

$$\begin{align*} \lim_{n \to \infty} \ln y &= \lim_{n \to \infty} \frac{\ln \left(1 - \frac{x}{n} \right)}{\frac{1}{n}} \\ &= \lim_{n \to \infty} \frac{\left(1 - \frac{x}{n}\right)^{-1} (x \cdot n^{-2})}{-n^{-2}} \\ &= \lim_{n \to \infty} -x\left(1 - \frac{x}{n} \right)^{-1} \\ &= -x \end{align*}$$

Therefore, since the exponential function $\exp(x) = e^x$ is continuous, we can move the limit in or outside of this function (by the definition of continuity) and thus find the limit of $y$ itself:

$$\lim_{n \to \infty} y = \lim_{n \to \infty} \exp(\ln y) = \exp \left( \lim_{n \to \infty} \ln y \right) = \exp (-x) = e^{-x}$$