 Secants straddling a differentiability point

If $u_n\le x\le v_n$, $u_n<v_n$ and both endpoints tend to $x$, differentiability gives $g(x+h)=g(x)+g^{\prime }(x)h+r(h)$ with $r(h)=o(|h|)$. The error in the secant is bounded by $\epsilon (|v_n-x|+|u_n-x|)/(v_n-u_n)=\epsilon$ for large $n$. Thus these secants converge even when either endpoint equals $x$.