I was trying to construct a proof of one form of L'Hopital's rule, and it seems correct to me. It would be nice to have a second opinion though. It'd be really nice if it's wrong someone could explain why.
$\textbf{Claim:}$Assume that $a\in\mathbb{R}$ and $f,g:(a,\infty)\to\mathbb{R}$ are differentiable. Moreover, suppose that $g'(x)\not=0$ and $\lim_{x\to\infty}g(x)=\infty$. If $$\lim_{x\to\infty}\dfrac{f'(x)}{g'(x)}\to\infty,$$ then $$\lim_{x\to\infty}\dfrac{f(x)}{g(x)}\to\infty.$$
$\textbf{Proof:}$ Suppose $R\in\mathbb{R}^+$. Notice that since $$\lim_{x\to\infty}\dfrac{f'(x)}{g'(x)}\to\infty$$ hence there exists a $P>a$ such that $(f'(x)/g'(x))>R$ for all $x\geq P$. Suppose that $P<y<x$. By Cauchy's M.V.T. it follows that $$R<\dfrac{f(x)-f(y)}{g(x)-g(y)}.$$ Choosing $x$ large enough we find $g(x)>0$ and $g(x)>g(y)$, so $[g(x)-g(y)]/g(x)>0$, which is possible as $g(x)\to\infty$ as $x\to\infty$. Then $$R\bigg(\dfrac{g(x)-g(y)}{g(x)}\bigg)<\dfrac{f(x)-f(y)}{g(x)-g(y)}\bigg(\dfrac{g(x)-g(y)}{g(x)}\bigg)=\dfrac{f(x)-f(y)}{g(x)}$$ $$R-R\bigg(\dfrac{g(y)}{g(x)}\bigg)+\dfrac{f(y)}{g(x)}<\dfrac{f(x)}{g(x)}.$$ As $g(y),R,f(y)$ are fixed and $g(x)\to\infty$ as $x\to\infty$ it follows that for all $\epsilon>0$ there exists a $Y>y$ such that if $x\geq Y$, then $$R-\epsilon<\dfrac{f(x)}{g(x)}.$$ It follows that we can apply this same process to $R'>R$, and setting $\epsilon=R'-R$ we can choose $Y'>y$ so that for all $x\geq Y'$ we find $$R<\dfrac{f(x)}{g(x)}.$$ Since $R$ was arbitrary greater than $0$ this shows that $f(y)/g(y)\to\infty$ as $y\to\infty$.$\square$
Your proof is fine, but the argument can be more concise.
For any $A,\epsilon > 0$ there exists $B_1 > 0$ such that $x > B_1$ implies $f'(x)/g'(x) > A + \epsilon$ and $g(x) > 0$.
Taking $B_1 < y < x$, by the MVT, there exists $\xi > y > B_1$ such that
$$\frac{\frac{f(x)}{g(x)}- \frac{f(y)}{g(x)}}{1 - \frac{g(y)}{g(x)}}= \frac{f(x) - f(y)}{g(x) - g(y)} = \frac{f'(\xi)}{g'(\xi)} > A + \epsilon,$$
which implies
$$\frac{f(x)}{g(x)} > A + \epsilon + \frac{f(y) - (A +\epsilon)g(y)}{g(x)}$$
With $y$ fixed , the last term on the RHS converges to $0$ as $x \to \infty$ and must be greater than $-\epsilon$ for $x > B_2$.
Thus, for all $x > \max(B_1,B_2)$ we have
$$\frac{f(x)}{g(x)} > A + \epsilon - \epsilon = A$$