$\displaystyle\lim_{x\to p} \ln(x) = \ln(p)$
Let $\epsilon>0$.
$|\ln(x)-\ln(p)| = \Big|\ln\Big(\frac{x}{p}\Big)\Big| $
Observe that:
$$\Big|\ln\Big(\frac{x}{p}\Big)\Big| < \epsilon \iff -\epsilon < \ln\Big(\frac{x}{p}\Big) < \epsilon \iff e^{-\epsilon} < \frac{x}{p} < e^\epsilon \iff pe^{-\epsilon} < x < pe^\epsilon \iff pe^{-\epsilon} - p < x - p < pe^\epsilon - p \iff p(e^{-\epsilon} - 1) < x - p < p(e^\epsilon-1)$$ $$ \iff -p(e^\epsilon - 1) < x - p < p(e^\epsilon - 1) \iff |x - p| < p(e^\epsilon - 1) $$
Note that $e^\epsilon - 1 \geq -e^{-\epsilon}+1$, since $e^x-1$ is convex, $-e^{-x}+1$ is concave and they share $y=x$ as tangent in $x=0$. Then $-(e^\epsilon - 1) \leq e^{-\epsilon}-1 $.
Then make $\delta = p(e^\epsilon - 1) > 0$.
So we have: $|x - p| < \delta \Rightarrow |\ln(x)-\ln(p)| < \epsilon$
I think my proof is ok, but I became confused after seeing what André Nicholas choosed for $\delta$ in this answer. Isn't this implying my choice for $\delta$ doesn't work?
Note that, $$\exp\left|\ln(x) - \ln(p )\right| = \max\{\tfrac{x}{p},\tfrac{p}{x}\}$$
The only real issue we get with $\ln$ is at values near $0$, so to make sure we're away from it, we require $\delta < p/2$, then $|x-p|<\delta$ implies $p/2 < x$. By the triangle inequality \begin{align} \frac{x}{p} &\leq \left|\frac{p}{p}\right| + \left|\frac{x-p}{p}\right| < 1 + \frac{\delta}{p} < e^{\delta/p}\\ \frac{p}{x} &\leq \left|\frac{x}{x}\right| + \left|\frac{p-x}{x}\right| < 1 + \frac{\delta}{(p/2)} < e^{2\delta/p} \end{align}
Hence $$\exp\left|\ln(x) - \ln(p )\right| < \exp(2\delta/p)$$so if $\delta = \min\{p/2, \epsilon p /2\}$, then $|\ln(x)-\ln(p )| < \epsilon$.