Can someone please unveil the steps for this answer?
Thus $$ \frac{f(x)}{g(x)}\approx \color{red}{\frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}}. $$ Taking the limit of the right hand side gives $\dfrac{f'(a)}{g'(a)}$.
Because $x \to a \iff x- a \to 0$, then
$$\lim_{x \to a} \color{red}{\dfrac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}} = \dfrac{f'(a)\times0 + f(a)}{g'(a)\times0 + g(a)}$$
Now what?
A smooth function is locally linear:
$$f(x)\approx f(a)+(x-a)f'(a)$$
and when $f(a)=0$, this simplifies to
$$f(x)\approx (x-a)f'(a).$$
L'Hospital easily follows:
$$f(a)=g(a)=0\implies \frac{f(x)}{g(x)}\approx \frac{f'(a)}{g'(a)}$$
and the approximation becomes more and more accurate as $x\to a$.
More rigorously,
$$f(x)=f(a)+(x-a)f'(a)+(x-a)r(x,a)$$ where $r$ tends to $0$ when $x$ tends to $a$.
Then
$$\frac{f(x)}{g(x)}=\frac{f'(a)+r(x,a)}{g'(a)+s(x,a)}$$ and
$$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(a)+r(x,a)}{g'(a)+s(x,a)}=\frac{f'(a)}{g'(a)}.$$
If $g(a)\ne0$, we can just write $\frac{f(x)}{g(x)}\approx\frac{f(a)}{g(a)}$.
If $f(a)\ne0=g(a)$, $\frac{f(x)}{g(x)}$ explodes to $\pm\infty$.
But if $f(a)=g(a)=0\ne g^\prime(a)$, that's more interesting.
In fact, your question contains almost all the logic we need for it: substituting $f(a)=g(a)=0,\,x\ne a$ gives$$\frac{f(x)}{g(x)}\approx\frac{f^\prime(a)(x-a)+0}{g^\prime(a)(x-a)+0}=\frac{f^\prime(a)(x-a)}{g^\prime(a)(x-a)}=\frac{f^\prime(a)}{g^\prime(a)}.$$
What you mention is an attempt to prove the following simple result :
Simple Theorem: If functions $f, g$ are differentiable at $a$ and $f(a)=g(a) =0,g'(a)\neq 0$ then $$\lim_{x\to a} \frac{f(x)} {g(x)} =\frac{f'(a)} {g'(a)} $$
The presentation in your question in not rigorous and the result is simply obtained using definition of derivative. We have $$\frac{f(x)} {g(x)} =\dfrac{\dfrac{f(x) - f(a)} {x-a}} {\dfrac{g(x) - g(a)} {x-a}} $$ and clearly as $x\to a$ the right hand side tends to $f'(a) /g'(a) $ via definition of derivative.
