Prove the following limit using the N-delta definition $\lim\limits_{x \to 0^+} {\ln x} = -\infty$
My attempt:
$\forall\ N<0,\ \exists \ \delta>0 \ st. \forall x,\
c < x < \delta + c \implies f(x) < N$
Let $N$ be given. 
Consider $\ \ln x < N$ 
$\; \; \; \; \; \; \; \;  \; \; \; \;  \;  e^{\ln x} < e^N$  
$\; \; \; \; \; \; \; \;  \; \; \; \;  \;  x < e^N$  
so I set $\delta = e^N$. 
The correct solution is $\delta = \frac{1}{e^m}$, and I'm not sure how they arrived at that.
Can anyone show me the way they arrive at that with a little explanation?

Your work is correct. But usually, the statement is:
$$\forall \ M > 0, \  \exists \ \delta > 0, \  0 < x < \delta \implies f(x) < -M. $$
That is, $M$ is taken positive. 
In this case, the choice would be $e^{-M}$, which is $1/e^{M}$.
Note that this is equivalent to what you have done:
You arrived at $\delta = e^N$, where $N < 0$. Now write that as $\delta = 1/e^{-N}$, and here $-N > 0$. That's the same thing.
