Limits of Lipschitz Continuous Composition of Functions

Question

Suppose $f:[\underline x, \overline x]\times \mathbb R \rightarrow \mathbb R$ is Lipschitz continuous with $f(\overline x,y)=0$ for all $y\in \mathbb R$, and suppose $g:[\underline x, \overline x] \rightarrow \mathbb R$ is continuously differentiable on $(\underline x, \overline x)$. Suppose $\lim_{x\rightarrow\overline x}(\overline x-x)g(x)=0$. It follows directly from Lipschitz continuity that $\lim_{x\rightarrow\overline x}f(x,y)g(x)=0$ for all $y\in \mathbb R$. I am trying to show that this implies $\lim_{x\rightarrow\overline x}f(x,g(x))g(x)=0$.

Answer 1

Since $f(\overline{x},g(x)) = 0$ we have \begin{align*} |f(x,g(x))g(x)| &\leq |f(x,g(x))g(x) - f(\overline{x},g(x))g(x)| + |f(\overline{x},g(x))g(x)|\\ &\leq |g(x)||f(x,g(x)) - f(\overline{x},g(x))|\leq L|g(x)||x-\overline{x}| \mathop{\longrightarrow}^{x\to \overline{x}^-} 0.\end{align*}