Prove the following two statements about differentiability are equivalent

Question

Let $E$ be a subset of $\textbf{R}$, $f:E\to\textbf{R}$ be a function, $x_{0}\in E$, and $L\in\textbf{R}$. Then the following two statements are equivalent

(a) $f$ is differentiable at $x_{0}$, and $f'(x_{0}) = L$

(b) We have $\displaystyle\lim_{x\to x_{0};x\in E-\{x_{0}\}}\frac{|f(x) - (f(x_{0}) + L(x-x_{0}))|}{|x-x_{0}|} = 0$

My solution

According to the definition of derivative, one has that \begin{align*} \lim_{x\to x_{0}}\frac{f(x) - f(x_{0})}{x-x_{0}} = L = \lim_{x\to x_{0}} L & \Leftrightarrow \lim_{x\rightarrow x_{0}}\left(\frac{f(x) - f(x_{0})}{x-x_{0}} - L\right) = 0\\\\ & \Leftrightarrow \lim_{x\rightarrow x_{0}}\frac{f(x) - (f(x_{0}) + L(x-x_{0}))}{x-x_{0}} = 0\\\\ & \Leftrightarrow \lim_{x\rightarrow x_{0}}\frac{|f(x) - (f(x_{0}) + L(x-x_{0}))|}{|x-x_{0}|} = 0 \end{align*}

Could someone tell me if it correct to make use of equivalences or should I break the proof into two implications?

Answer 1

There are many equivalent ways to define differentiability. If (as I suppose) your definition of "$f$ is differentiable at $x_0$" is that the limit $$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x -x_0}$$ exists and is denoted by $f'(x_0)$, then yes, your proof is correct.

Note that in your last equivalence you are implicitly using two (obvious) facts:

1. for any $a \in \mathbb{R}$, if $|a| = 0$ then $a = 0$;
2. $\lim_{x \to x_0} |g(x)| = |\lim_{x \to x_0} g(x)|$.