I am trying to rigorously prove a result about derivatives when removable discontinuities are involved. Specifically, I have a bi-variate function $f(a,b)$ that is undefined for $a=0$. However, $\forall b\in\mathbb{R}$ the limit $$\lim_{a\rightarrow 0} f(a,b)$$ is well-defined and converges to a given constant $f_0$. I therefore eliminate the discontinuity by considering $$ F(a,b)= \begin{cases} f(a,b) & a\neq 0 \\ f_0 & a= 0 \end{cases} $$
I now consider two additional functions $x(t)$ and $y(x,t)$. The latter is bounded everywhere, and continuous for all $x\neq 0$. In $x=0$, it has a removable discontinuity. These functions are passed as inputs of $F$, thus generating a further function $G(t)=F\left(x(t),y(t)\right)$. I was able to compute the time derivative of $G$ using the chain rule and assuming $x\neq 0$. $G^\prime(t)$ writes as a sum of different components: $$ G^\prime = h(x)\,y + m(t) $$ with $h(x)$ being undefined for $x=0$. However, I can show that $\lim_{x\rightarrow 0} h(x)$ exists and equals 0.
The question is: can I say that the derivative of $G$ is
$$ G^\prime = H(x)\,y + m(t) $$
with
$$
H(x)=
\begin{cases}
h(x) & x\neq 0 \\
0 & x = 0
\end{cases}
$$
My intuition says yes, but I am not sure how to prove it. The idea I have is to first consider a corollary of De L'Hopital's theorem: if a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous in a point $x_0$ and $\lim_{x\rightarrow x_0} f^\prime(x)$ exists finite, then $f$ is differentiable at $x_0$ and $f^\prime(x_0) = \lim_{x\rightarrow x_0} f^\prime(x)$. Applied to my case: $G$ is continuous everywhere, and for any $t_0 : x(t_0) = 0$ then: $$\lim_{t\rightarrow t_0}G^\prime = \lim_{t\rightarrow t_0} h(x(t))\,y(x(t),t) + m(t) = \lim_{x\rightarrow 0} h(x)\,y(x,t_0) + m(t_0) = m(t_0)$$ The corollary should therefore hold... or not? Could you find any mistake or confirm my reasoning? Thanks in advance for any help!