How to find $\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n$ 
Let $c \neq 1$ be a postive real number. Find the following limit
$$\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n.$$

I know that $\lim_{n \to \infty} \Big( 1 + \frac{c}{n} \Big)^{bn} = e^{bc}$. But it cannot be used here, since $\ln(n)$ is not a constant.  I also know how to find the $\lim_{n \to \infty} \Big( 1 + \frac{1}{n^2} \Big)^{n}$, which is maybe closer to our limit.
I tried to use some methods (e.g., L'Hopital's rule) that appear in the proofs of the above limits to solve also $\lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n$ but without success. The graphs of $\Big (1 - \frac{c \ln(n)}{n} \Big)^n$ indicate that the limit is $\infty$ if $c < 1$ and the limit is 0 otherwise.
I appreciate any suggestions on how to solve this limit. Also, any readings on this topic (I did not find any such example anywhere) are appreciated.
 A: $\frac {\ln (1-x)} x \to -1$ as $x \to 0$. To find the limit of $n \ln (1-\frac {c \ln n} n)$ it is therefore enough to find the limit of $-n \frac {c \ln n} n$ which is clearly $-\infty$ for any $c>0$ and $\infty$ for $c <0$. Now take exponential.
Answer: The limit is $0$ if $c>0$, $\infty$ if $c <0$ and $1$ if $c=0$.
A: If $x_n\to \infty$, and $c>0$, then
$$
\left(1-\frac{c}{x_n}\right)^{x_n}\to \mathrm{e}^{-c}
$$
In the OP case
$$
\left(1-\frac{c\ln n}{n}\right)^{n}=
\left(1-\frac{c}{\frac{n}{\ln n}}\right)^{\frac{n}{\ln n}\cdot \ln n}=
\left(\left(1-\frac{c}{x_n}\right)^{x_n}\right)^{\ln n}
$$
where $x_n=\frac{n}{\ln n}$. Hence
$$
\lim_{n\to\infty}\left(1-\frac{c\ln n}{n}\right)^{n}=\lim_{n\to\infty}\left(\mathrm{e}^{-c}\right)^{\ln n}=\lim_{n\to\infty}n^{-c}=0.
$$
A: Since $$\lim_{n \to \infty}(1 - \frac{c \ln(n)}{n}) = 1 $$We can write the limit as $$L = \lim_{n \to \infty} \Big (1 - \frac{c \ln(n)}{n} \Big)^n = \lim_{n \to \infty} \exp(-\frac{c \ln(n)}{n})n = \lim_{n\to \infty} \exp(-c \ln(n)) = \lim_{n \to \infty} n^{-c}$$If $c \gt 0$ then $L = 0$. In the case $c\lt 0$ we have $L = \infty$ and $c = 0$ implies $L = 1$.
A: Using the Maclaurin series of the logarithm and the exponential function, we have
\begin{align*}
\left( {1 - \frac{{c\log n}}{n}} \right)^n & = \exp \left( {n\log \left( {1 - \frac{{c\log n}}{n}} \right)} \right) = \exp \left( {n\left( { - \frac{{c\log n}}{n} + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{{n^2 }}} \right)} \right)} \right) \\ &= \exp \left( { - c\log n + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right)} \right) = n^{ - c} \left( {1 + \mathcal{O}\!\left( {\frac{{\log ^2 n}}{n}} \right)} \right).
\end{align*}
It is now easy to conclude.
