Can a limit of form$\ \frac{0}{0}$ be rational if the numerator is the difference of transcendental functions, and the denominator a polynomial one?

Question

Let$\ f_1(x)$ and$\ f_2(x)$ be transcendental functions such that$\ \lim_{x\to 0} f_1(x)-f_2(x)=0$, and$\ f_3(x) $ polynomial, such that$\ f_3(0)=0$. Can$\ \lim_{x\to 0} \frac{f_1(x)-f_2(x)}{f_3(x)}$ be rational?

Answer 1

If $f_3(0)\ne0$ then $$ \lim_{x\to 0} \frac{f_1(x)-f_2(x)}{f_3(x)}=0\in\mathbb{Q}. $$ The above example was given before the OP added the condition $f_3(0)=0$.

If you want something less trivial $$ \lim_{x\to0}\frac{e^{2x}-e^x}{x}=1\in\mathbb{Q}. $$

Answer 2

A more boring example: Let $f$ be a transcendental function. Then $$\frac{(f(x)+x)-f(x)}{x}$$ works.

Answer 3

$$ \frac{\sin x -(e^x-1)}{x^2} \to - \frac12 $$