Given that: $$\lim_{x \to c}f(x)=L_1$$which means(1): $$\forall\epsilon_1>0, \exists \delta_1>0:|x-c|<\delta_1\implies|f(x)-L_1|<\epsilon_1 $$ and $$\lim_{x \to c}g(x)=L_2, \space \text{where}\space g(x), \space L_2\ne0 $$ which means(2): $$\forall\epsilon_2>0, \exists \delta_2>0:|x-c|<\delta_2\implies|g(x)-L_2|<\epsilon_2 $$
I want to prove: $$\lim_{x \to c}\frac{f(x)}{g(x)}=\frac{\lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}=\frac{L_1}{L_2}$$
which means showing: $$\forall \epsilon>0,\exists \delta>0:|x-c|<\delta\implies|\frac{f(x)}{g(x)}-\frac{L_1}{L_2}|<\epsilon$$
First, let: $$\delta=\min(\delta_1,\delta_2),\space \epsilon=2a, \space \epsilon_1=ak,\space \epsilon_2=an$$where $$ \frac{1}{k}>\frac{1}{|L_2|} \space \text{and} \space \frac{1}{n}>|\frac{f(x)}{g(x)L_2}|$$ $k,n>0$ from how we defined them.
Substituting $\epsilon_1$ in (1) gives us:$$|f(x)-L_1|<\epsilon_1=ak$$ which implies(3): $$\frac{1}{|L_2|}\cdot |f(x)-L_1|<a$$
The same procedure is done for (2), which gives: $$|g(x)-L_2|<\epsilon_2=an$$ which implies(4): $$|\frac{f(x)}{g(x)L_2}|\cdot|g(x)-L_2|<a$$
Adding (3) and (4) gives:$$\frac{1}{|L_2|}\cdot |f(x)-L_1|+|\frac{f(x)}{g(x)L_2}|\cdot|g(x)-L_2|=\frac{1}{|L_2|}\cdot |f(x)-L_1|+|\frac{f(x)}{g(x)L_2}|\cdot|L_2-g(x)|\\=|\frac{f(x)-L_1}{L_2}|+|\frac{f(x)}{g(x)}-\frac{f(x)}{L_2}|\ge |\frac{f(x)-L_1}{L_2}+\frac{f(x)}{g(x)}-\frac{f(x)}{L_2}|$$
Finally, $$|\frac{f(x)-L_1}{L_2}+\frac{f(x)}{g(x)}-\frac{f(x)}{L_2}|=|-\frac{L_1}{L_2}+\frac{f(x)}{g(x)}|=|\frac{f(x)}{g(x)}-\frac{L_1}{L_2}|<2a=\epsilon$$ and $|\frac{f(x)}{g(x)}-\frac{L_1}{L_2}|$ is indeed less than $\epsilon$. We have proved what we wanted.
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$\begingroup$Your proof is fine. If you have the limit theorem for products, you can instead apply that to $f(x)$ and $1/g(x)$ rather than having to go through all of that (but you would have to also prove $\lim 1/g(x) = 1/\lim g(x)$ which is essentially the difficulty).
$\endgroup$ $\begingroup$The proof is correct except for a minor step. You propose that there is a number $n>0$ such that $|f(x) /g(x) L_{2}|<1/n$. This is true but not obvious / self-evident. Note that by choosing $\epsilon=1$ in the limit definition for $f$ we have a $\delta_{3}>0$ such that $$0<|x-a|<\delta_{3}\implies |f(x) - L_{1}|<1\implies |f(x) |<|L_{1}|+1$$ and choosing $\epsilon=|L_{2}|/2>0$ in the limit definition of $g$ we have a $\delta_{4}>0$ such that $$0<|x-a|<\delta_{4}\implies|g(x)-L_{2}|<|L_{2}|/2\implies |g(x) |>|L_{2}|/2$$ Therefore $$0<|x-a|<\min(\delta_{3},\delta_{4})\implies \left|\frac{f(x)} {g(x) L_{2}}\right|<\frac{2(|L_{1}|+1)}{|L_{2}|^{2}}=\frac{1}{n}\text{ (say)} $$ Now on the basis of $\epsilon, n, k$ choose your $\delta_{1},\delta_{2}$ and keep $\delta$ as minimum of all the deltas.
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