I'm studying for an exam in an electrical engineering course (stochastic process in dynamic systems), though this section is strictly on the math. A given practice problem (with no solution given, of course) is:
"Show that if $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is continuously differentiable around $x_0$, then $$f(x) = f(x_0) + A(x-x_0)+g(x)$$ where $$\lim_{x\to x_0} \frac{||g(x)||}{||x-x_0||}=0$$
Note that in the first equation, A is the matrix of partial derivatives of $f$ at $x_0$
The approach I've taken, which I don't think is correct, is as follows:
$f$ is continuously differentiable, then $$\lim_{x\to x_0}\frac{f(x)-f(x_0)-A\dot(x-x_0)}{||x-x_0||}=0$$
So we have $$\lim_{x\to x_0}\frac{f(x)-f(x_0)-A\dot(x-x_0)}{||x-x_0||} = \lim_{x\to x_0} \frac{||g(x)||}{||x-x_0||}=0$$
$$\frac{\lim_{x\to x_0}f(x)-f(x_0)-A\dot(x-x_0)}{\lim_{x\to x_0}||x-x_0||} = \frac{\lim_{x\to x_0} ||g(x)||}{\lim_{x\to x_0} ||x-x_0||}$$
$$\lim_{x\to x_0}f(x)-f(x_0)-A\dot(x-x_0) = \lim_{x\to x_0} ||g(x)||$$
$$\lim_{x\to x_0}f(x) = \lim_{x\to x_0} f(x_0)+A\dot(x-x_0)+||g(x)||$$
Since the codomain of $f(x)$ is $\mathbb{R}$, we can assume the codomain of $g(x)$ is also $\mathbb{R}$, meaning $||g(x)|| = \sqrt{g(x)^2} = g(x)$, which leaves us with
$$\lim_{x\to x_0}f(x) = \lim_{x\to x_0} f(x_0)+A\dot(x-x_0)+g(x)$$
Here, I'm tempted to just drop the limits, leaving the equation which was to be shown. But I feel like this is wrong, because it seems strange to equate two functions because their limits are the same. For example, if $r(x)=x$ and $s(x)=\sin(x)$, $$\lim_{x\to 0}r(x) = 0 = \lim_{x\to 0}s(x)$$
But certainly $x\neq \sin(x)$
Can anyone help me solve this?