Show $\lim\limits_{n\to\infty} n \ln\left(1-\frac{1}n\right) = -1$ Could you help me show that $$\lim\limits_{n\to\infty} n \ln \left({1-\frac{1}n} \right) = -1 ?$$
 A: Another way, and the way I would do it in practice:
We have the power series expansion
$\log(1-1/x)=-\frac{1}{x}+O\left(\frac{1}{x}^2\right)\dots$
So
$x\log(1-1/x)=-1+O\left(\frac{1}{x}\right)$
and letting $x$ tend to infinity gives the answer. 
A: Recall that for $x>0$, $$1-\frac{1}{x} \leq \log x \leq x-1.$$ Then $$\frac{-n}{n-1} \leq n \log \left(1-\frac{1}{n}\right) \leq -1,$$ and the result follows by taking $n \rightarrow \infty$.
The first inequality can be derived from $\log x= \int_1^x \frac{dt}{t}$, since $$\int_1^x \frac{dt}{t^2} \leq \int_1^x \frac{dt}{t} \leq \int_1^x dt.$$
A: $$\lim\limits_{n\to\infty} n \ln\left({1-\frac{1}n}\right)=\lim_{n\to\infty}  \ln\left({1-\frac{1}n}\right)^n=\ln\lim_{n\to\infty}\left(1-{1\over n}\right)^n=\ln e^{-1}=-1$$
A: We take the limit:
$\lim\limits_{n\to\infty} n \ln\left({1-\frac{1}n}\right)$
This is a limit of the type '0 $\infty$'. Let $t=\frac{1}n$:
$=\lim\limits_{t\to 0} \dfrac{\ln\left({1-t}\right)}t$
This is a limit of the type '0/0'. We apply L'Hospital's rule:
$=\lim\limits_{t\to 0} \dfrac{1}{t-1}$
The limit of the quotient is defined if it is defined for numerator and denominator. The limit of the constant '1' in the numerator is the constant '1'.
$=\dfrac{1}{\lim\limits_{t\to 0} t-1}$
The answer is therefore
$= -1$
(Solved with the help of wolfram alpha)
A: Using L'Hopital rule, $$ \lim\limits_{n\to\infty} n \ln \left({1-\frac{1}n} \right) = \lim_{n \rightarrow \infty} \frac{ \ln \left(1 - \frac 1 n \right)}{ \frac 1 n} = \lim_{n \rightarrow \infty} \frac{\frac{1}{1 - \frac 1 n}}{- \frac 1 {n^2}} \times \frac{1}{n^2} = -1$$
A: Recall that $$\lim_{n \to \infty}\left( 1 - \dfrac1n \right)^n = e^{-1}$$
Since $\log$ is a continuous function, we have that
$$\lim_{n \to \infty} n \log\left( 1 - \dfrac1n \right) = \lim_{n \to \infty} \log\left( 1 - \dfrac1n \right)^n = \log \left( \lim_{n \to \infty}\left( 1 - \dfrac1n \right)^n \right) = \log \left(e^{-1} \right) = -1$$
