Suppose that $f(0) = g(0) = 0$ for two differentiable functions $f$ and $g$. The limit $$\lim_{x\to 0}{\dfrac{f}{g}}$$ is equal to:
a) $\lim_{x\to 0}{\dfrac{f'}{g'}}$.
b) $\lim_{x\to 0}{\dfrac{f'g-fg'}{g^2}}$.
c) $\lim_{x\to 0}{\dfrac{fg'-f'g}{g^2}}$.
d) $\lim_{x\to 0}{\dfrac{f'}{g}}$.
e) $\lim_{x\to 0}{\dfrac{f}{g'}}$.
Any guidance is appreciated since I'm not really sure how to approach this problem
If both $f$ and $g$ are differentiable at $0$, then they are continuous at $0$, so $\lim\limits_0f=\lim\limits_0g=0$ (from $f(0)=g(0)=0$). Now, we can use L'Hospital's rule:
$$\text{If }\,\, \exists \lim\limits_0\frac{f'}{g'}\,\,\text{, then }\,\,\exists \lim\limits_0\frac{f}{g}\,\, \text{ and }\,\,\lim\limits_0\frac{f}{g}=\lim\limits_0\frac{f'}{g'}$$ L'Hospital's rule: Let $-\infty\leq a<b\leq\infty$ and $f\text{,}g:]a,b[ \to \mathbb{R}$, $g(x)\neq0 \,(x \in]a,b[)$, $\exists\lim\limits_bf=\lim\limits_bg=0$. If $f$ and $g$ are differentaible and $\exists \lim\limits_b\frac{f'}{g'}$, then $\exists \lim\limits_b\frac{f}{g}=\lim\limits_b\frac{f'}{g'}$
Edit: $$\lim_0\frac{f}{g}=\lim_0\frac{f(x)}{g(x)}\frac{1/x}{1/x}$$ $$=\lim_0\frac{f(x)/x}{g(x)/x}$$ Now, we can subtract $0$ from 'everywhere', that won't change anything: $$=\lim_0\frac{\frac{f(x)-0}{x-0}}{\frac{g(x)-0}{x-0}}$$ Now we can take the limit of the top and bottom separately and use the fact that $f(0)=g(0)=0$: $$=\frac{\lim\limits_0\frac{f(x)-f(0)}{x-0}}{\lim\limits_0\frac{g(x)-g(0)}{x-0}}$$ We can see that the top is the definition of $f'(0)$, and the bottom is the definition of $g'(0)$: $$=\frac{f'(0)}{g'(0)}$$ And if $f'$ and $g'$ are continuous we have: $$=\lim_0\frac{f'}{g'}$$ However, we were using some rules which are not always true (for example $\lim \frac{f}{g}$ is not always $\frac{\lim f}{\lim g}$). But if we assume that 'everything is nice', then the A is the correct answer.