= Uniform derivative convergence with an anchored value
Let $f_n$ be <continuously differentiable functions> on an <interval> $I$. If $f_n'\to g$ by <uniform convergence> on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a <continuously differentiable function> $f$ with $f'=g$. If $I$ is bounded, the convergence is uniform on all of $I$.<Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl/6.2 Interchange of limits>[Basic Analysis: Introduction to Real Analysis, Volume I by Jiří Lebl, §6.2.3, Theorem 6.2.10 and the unbounded-interval observation following its proof]{ref}
The anchor fixes the additive constant that a <derivative> cannot see. By the <fundamental theorem of calculus>, define
$$
f(x)=A+\int_c^xg(t)\,dt,\qquad |f_n(x)-f(x)|\leq|f_n(c)-A|+|x-c|\|f_n'-g\|_\infty.
$$
The <uniform limit theorem> makes $g$ a <continuous function>, so the <fundamental theorem of calculus> gives $f'=g$. The displayed estimate proves <uniform convergence> wherever $|x-c|$ has a common finite bound, and therefore proves <locally uniform convergence> on any <interval>.
For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its <derivative> is $x+1/n$, with <supremum error> $1/n$ from $g(x)=x$, and $f_n(0)\to3$. The reconstructed <uniform limit> is $f(x)=3+x^2/2$; the exact <supremum error> is $3/n$, equal to the bound above. Without an anchor, the constant <functions> $f_n(x)=n$ have zero <derivatives> but no finite <pointwise limit>. On the <real line>, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent <derivatives>, but its <supremum error> from zero is infinite. The bounded-distance condition has a real role.
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