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:
 $$\lim_{n\to\infty}\left(\frac{n-x}{n}\right)^n=\frac{1}{e^x}$$
 A: *

*(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 
$$\begin{eqnarray*}
\lim_{n\rightarrow \infty }\left( \frac{n-1}{n}\right) ^{n}
&=&\lim_{n\rightarrow \infty }\frac{1}{\left( \frac{n}{n-1}\right) ^{n}}=
\frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n}{n-1}\right) ^{n}} \\
&=&\frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right) ^{n+1}}=
\frac{1}{\lim_{n\rightarrow \infty }\left( \frac{n+1}{n}\right)
^{n}\cdot\lim_{n\rightarrow \infty }\frac{n+1}{n}} \\
&=&\frac{1}{\mathrm{e}\cdot 1}=\frac{1}{\mathrm{e}}.
\end{eqnarray*}$$

*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.

A: 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.
A: 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:
$$\lim_{n \to \infty} y = \lim_{n \to \infty} \exp(\ln y) = \exp \left( \lim_{n \to \infty} \ln y \right) = \exp (-x) = e^{-x}$$
A: I do the first part and the second part is done similarly.
Solution 1 We know that the derivative of $\ln x$ is $\frac{1}{x}$. Hence using first principle differentiation, we have:
$$
\begin{align*}
\frac{1}{x}&=\lim_{h\to0}\frac{\ln(x+h)-\ln x}{h}\\
&=\lim_{h\to0}\ln\left(1+\frac{h}{x}\right)^{\frac{1}{h}}
\end{align*}
$$
Let $n=\frac{1}{h}$ and $y=\frac{1}{x}$, so now
$$
y=\lim_{n\to\infty}\ln\left(1+\frac{y}{n}\right)^n
$$
Now let $y=-1$, then
$$
\begin{align*}
-1&=\lim_{n\to\infty}\ln\left(1-\frac{1}{n}\right)^n\\
\frac{1}{e}&=\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n
\end{align*}
$$
Solution 2 Use L'Hopital's rule:
$$
\begin{align*}
\lim_{n\to\infty}n\ln \left(1-\frac{1}{n}\right)&=-\lim_{n\to\infty}\frac{1}{1-\frac{1}{n}}\\
&=-1
\end{align*}
$$
The rest is like solution 1.
A: For the first, take the reciprocal and substitute $n\mapsto n+1$ to get
$$
\begin{align}
\frac{1}{\lim\limits_{n\to\infty}\left(\frac{n-1}{n}\right)^n}
&=\lim_{n\to\infty}\left(\frac{n}{n-1}\right)^n\\
&\stackrel{n\to n+1}{=}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^{n+1}\\
&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\\
&=e\cdot1
\end{align}
$$
Therefore, $\lim\limits_{n\to\infty}\left(\frac{n-1}{n}\right)^n=\dfrac1e$.
For the second, take the reciprocal and substitute $n\mapsto nx+x$ to get
$$
\begin{align}
\frac{1}{\lim\limits_{n\to\infty}\left(\frac{n-x}{n}\right)^n}
&=\lim_{n\to\infty}\left(\frac{n}{n-x}\right)^n\\
&\stackrel{n\to nx+x}{=}\lim_{n\to\infty}\left(\frac{nx+x}{nx}\right)^{nx+x}\\
&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\cdot\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^x\\
&=e^x\cdot1
\end{align}
$$
Therefore, $\lim\limits_{n\to\infty}\left(\frac{n-x}{n}\right)^n=\dfrac{1}{e^x}$.
We could have started from the second and set $x=1$ for the first.
