My initial (incorrect) attempt:
$\lim_{n \to \infty} \ln(n+4) - \ln(n)$ = $\lim_{n \to \infty} \frac{\ln(n+4)}{\ln(n)}$
Since $\lim_{n \to \infty} \ln(n+4) = \infty$ and $\lim_{n \to \infty} \ln(n) = \infty$ then $\lim_{n \to \infty} \frac{\ln(n+4)}{\ln(n)}=\frac{\infty}{\infty}$; an indeterminate form.
Using L'Hospital's Rule: $\lim_{n \to \infty} \frac{\ln(n+4)}{\ln(n)}$ becomes $\lim_{n \to \infty} \frac{\frac{1}{n+4}}{\frac{1}{n}}$ or $\lim_{n \to \infty} \frac{n}{n+4}$.
Since $\lim_{n \to \infty} n = \infty$ and $\lim_{n \to \infty} n+4 = \infty$ then $\lim_{n \to \infty} \frac{n}{n+4}=\frac{\infty}{\infty}$; also an indeterminate form.
Using L'Hospital's Rule again: $\lim_{n \to \infty} \frac{n}{n+4}$ becomes $\lim_{n \to \infty} \frac{1}{1}$ or $\lim_{n \to \infty} 1=1$.
Where did I go wrong?
$$\ln {(n+4)}- \ln {(n)}=\underbrace{\ln {\Bigg(\frac{n+4}{n}\Bigg)}}_{\text{Correct}} \neq \underbrace{\frac{\ln(n+4)}{\ln(n)}}_{\text{What you did}}$$
$$\text{Since}~:\lim_{n \to \infty} \Bigg(\frac{n+4}{n}\Bigg)=1 \implies \lim_{n \to \infty} \ln \Bigg (\frac{n+4}{n}\Bigg)= \ln 1 =0 $$