Let $f: \mathbb{R}^{>0} \times \mathbb{R} \to \mathbb{R}^{\geq 0}$. Let $g$ be a function defined on $D := (-\infty, -1) \cup \{0\} \cup (1, \infty)$.
Suppose that for all $u \in (0, \infty) \bigcap \left(\{0\} \cup (1, \infty) \right)$ we have $f(u, g(u)) = 16u^2$.
Is it true that $\lim_{u \to 0} f(u, g(u)) = 0$?
[Original statement of this question follows below.]
I have a function $f$ defined on $(0;\infty[\times\Bbb{R}\to \Bbb{R}^+: (u,v)\to f(u,v)$. Let $g$ be a function defined on $D=]-\infty;-1)\cup \{0\}\cup (1;\infty[\to \Bbb{R}:v\to g(v)$
For $\forall u\in(0;\infty[ \cap\Big( \{0\}\cup (1;\infty[ \Big)$ we have $f\big(u,g(u)\big)=16u^2$.
My question can we say that $\lim_{u\to 0}f\big(u,g(u)\big)=0$
Note that $(0;\infty[:=\{x\in\Bbb{R}, x>0\}$ and $\Bbb{R}^+:=\{x\in\Bbb{R}, x\geq 0\}$
Note that the condition is equivalent to: for all $u > 1$ we have $f(u, g(u)) = 16u^2$.
Since we have no information about the behaviour of $z \mapsto f(z, g(z))$ near $z=0$, we can't say anything about its limit there.