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.

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.
