Finding a limit converging into $\ln(x)$

It's probably better to think to $$ \ln x=\int_{1}^{x} \frac{1}{t}\,dt $$ Since, for $a\ne -1$, $$ \int_{1}^{x} t^a\,dt=\frac{x^{a+1}-1}{a+1} $$ we can conjecture that $$ \lim_{a\to-1}\frac{x^{a+1}-1}{a+1}=\ln x $$ which is indeed true, because this is the derivative at $-1$ of the function $$ f(a)=x^{a+1} $$ and $$ f'(a)=x^{a+1}\ln x $$

When dealing with “indefinite integrals” it's common to get wrong conclusions, because of the constant of integration.