$\begingroup$ The logarithm is a function, meaning that it has actually a well defined value for a offered $x$. Girlfriend can't leave an undetermined continuous in the meaning ! $\endgroup$
Because that is not true that us have$$\log(1+x)=x-\fracx^22+\fracx^33-\fracx^44+\cdots+C$$for an arbitrary consistent $C$. Since, once $x=0$, the LHS is $0$ and also RHS is $C$, $C=0$.
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Since the original role is $\log (1+x)$ and for $x=0$ we have actually $\log (1+0)=0$ we require that likewise the collection is zero for $x=0$.
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