Proof

By the definition of limit:
- there is a $\delta_1 > 0$ such that if $x\in D$ and $0 < |x-a| < \delta_1,$ then $|g(x)-H| < \frac {|H|}{2}.$ For these $x$ it follows from the triangle inequality that $$\begin{array}{rcl}|g(x)| + \frac{|H|}2 & > & |g(x)|+|g(x)-H|\\ &=&|g(x)|+|H-g(x)|\\ &\ge &|g(x)+(H-g(x))|\\ &=&|H|,\end{array}$$ which implies $|g(x)| > \frac{|H|}2,$ and
- there is a $\delta_2 > 0$ such that if $x\in D$ and $0 < |x-a| < \delta_2,$ then $|f(x)-L| < \frac {\epsilon\cdot |H|}{4}.$
- Moreover, there is a $\delta_3 > 0$ sucht that if $x\in D$ and $0 < |x-a| < \delta_3,$ then $|g(x)-H| < \frac{\epsilon\cdot H^2}{4(|L|+1)}.$
Get the minimum $\delta=\min(\delta_1,\delta_2,\delta_3).$
Note that $\delta > 0.$
Since $\frac fg:D\to\mathbb R,$ we have that if $x\in D$ with $0 < |x-a| < \delta,$ then by the triangle inequality. $$\begin{array}{rcl}\left|\frac{f(x)}{g(x)}-\frac LH\right|&=&\left|\frac{f(x)\cdot H-L\cdot g(x)}{g(x)\cdot H}\right|\\ &=&\left|\frac{f(x)\cdot H-L\cdot H+L\cdot H-L\cdot g(x)}{g(x)\cdot H}\right|\\ &=&\left|\frac{(f(x)-L)\cdot H+L\cdot(H-g(x))}{g(x)\cdot H}\right|\\ &\le&\frac{|(f(x)-L)}{|g(x)|}+\left|\frac{L(g(x)-H)}{g(x)\cdot H}\right|\\ & < & \frac{e\cdot |H|}{4}\cdot \frac{2}{|H|} + \frac{\epsilon\cdot H^2}{4(|L|+1)}\cdot\frac{2|L|}{H^2}\\ & < & \frac \epsilon2 + \frac \epsilon2\\ &=&\epsilon\end{array}.$$