So we all know that, if $f$ and $g$ are defined on $[a,b]$ and are differentiable at a point $x \in [a,b]$, then $f/g$ is differentiable at $x$ if $g(x) \neq 0$. And the quotient rule says that $\left (\frac{f}{g} \right)'(x) = \frac{g(x) f'(x) - g'(x) f(x)}{g^2(x)}$.
Rudin's proof of Theorem 5.3 on page 105 says that, let $h = f/g$, then\begin{equation}\lim_{t \to x} \frac{h(t)-h(x)}{t-x} = \lim_{t \to x} \frac{1}{g(t)g(x)} \left[ g(x) \frac{f(t)-f(x)}{t-x} - f(x) \frac{g(t)-g(x)}{t-x} \right] = \frac{g(x) f'(x) - g'(x) f(x)}{g^2(x)}.\end{equation}
I understand why the algebra here is correct, but I can't see why my work below is wrong (although it certainly is). I have\begin{equation}\varphi(t) = \frac{\frac{f(t)}{g(t)}-\frac{f(x)}{g(x)}}{t-x} = \frac{f(t)}{g(t) \cdot (t-x)} - \frac{f(x)}{g(x) \cdot (t-x)}.\end{equation}Then,\begin{align}\lim_{t \to x}\varphi(t) & = \lim_{t \to x} \left[ \frac{f(t)}{g(t) \cdot (t-x)} - \frac{f(x)}{g(x) \cdot (t-x)} \right]\\& = \lim_{t \to x} \frac{f(t)}{g(t) \cdot (t-x)} - \lim_{t \to x} \frac{f(x)}{g(x) \cdot (t-x)}\\& = \lim_{t \to x} \frac{1}{g(t)} \cdot \lim_{t \to x} \frac{f(t)}{t-x} - \lim_{t \to x} \frac{1}{g(x)} \cdot \lim_{t \to x} \frac{f(x)}{t-x}\\& = \frac{1}{g(x)} \cdot \lim_{t \to x} \frac{f(t)}{t-x} - \frac{1}{g(x)} \cdot \lim_{t \to x} \frac{f(x)}{t-x}\\& = \frac{1}{g(x)} \cdot \left( \lim_{t \to x} \frac{f(t)}{t-x} - \lim_{t \to x} \frac{f(x)}{t-x} \right)\\& = \frac{1}{g(x)} \cdot \left( \lim_{t \to x} \frac{f(t) - f(x)}{t-x} \right)\\& = \frac{f'(x)}{g(x)}.\end{align}
I believe there must be some algebraic errors during the computation, but I can't see where. Thanks for any help!