5. $\lim _{x \rightarrow 2} \frac{\ln \left(x^{2}-3\right)}{x-2}=$
5. 4 Detailed Explanation: Let $x-2=t$, then $\lim _{x \rightarrow 2} \frac{\ln \left(x^{2}-3\right)}{x-2}=\lim _{t \rightarrow 0} \frac{\ln \left(t^{2}+4 t+1\right)}{t}$ $=\lim _{t \rightarrow 0} \frac{\ln \left(t^{2}+4 t+1\right)-\ln 1}{t-0}$, which is the derivative of $f(t)=\ln \left(t^{2}+4 t+1\right)$ at point 0. Also, $\left(\ln \left(t^{2}+4 t+1\right)\right)^{\prime}=\frac{2 t+4}{t^{2}+4 t+1}$, so when $t=0$, the value is 4.