Epsilon Delta Proof of Differentiability implies Continuity

Question

Can someone check my proof if a function $f:{R\mapsto R}$ is differentiable at all points, it is continuous throughout as well?

Proof:

Let $x$ be a real number, since $f(x)$ is differentiable, or $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists and is equal to $f'(x)$. So, for any $\epsilon > 0$ there exists a $\delta > 0$, such that $0<|h - 0| < \delta$

$\implies |\frac{f(x+h)-f(x)}{h} - f'(x)| < \epsilon$

$\implies |\frac{f(x+h)-f(x)-f'(x).h}{h}| < \epsilon$

$\implies |f(x+h)-f(x)-f'(x).h| < |h|\epsilon$

$\implies -|h|\epsilon<f(x+h)-f(x)-f'(x).h < |h|\epsilon$

$\implies f'(x).h-|h|\epsilon<f(x+h)-f(x) < f'(x).h+|h|\epsilon$

Let $\epsilon'$ be min($f'(x).h+|h|\epsilon - f(x+h)+f(x), f'(x).h-|h|\epsilon-f(x+h)+f(x) $), then we can say,

$\implies -\epsilon'<f(x+h)-f(x) < \epsilon'$

$\implies |f(x+h)-f(x)| < \epsilon'$

$\implies \lim_{h \to 0}f(x+h)$ exists and is equal to $f(x)$

$\implies f$ is continuous at $x$

NOTE: This is not a homework question, I am trying to learn basic calculus on my own through the web.

Answer 1

This is not correct. You cannot choose $\varepsilon'$. What you are supposed to prove is that for every $\varepsilon'>0$, there is a $\delta>0$ such that$$|h|<\delta\implies\bigl|f(x+h)-f(x)\bigr|<\varepsilon'.$$

Answer 2

Your proof does not work. It seems you have a misunderstanding of the definition of continuity. Your mistake is that your $\epsilon'$ depends on $h$. But this makes no sense (why? hint: continuity means that for every arbitrary margin of error $\epsilon$ you can find a neighborhood of $h$ where your function will be within that margin of error)

Try to fix it now that you understand and check the accepted answer here:

How to prove differentiability implies continuity with $\epsilon-\delta$ definition?

Answer 3

Simple proof without explicit use of $\epsilon$-$\delta$:

For any $x$ we have $$ \left| f(x+h) - f(x) \right| = \left| \frac{f(x+h) - f(x)}{h} h \right| = \left| \frac{f(x+h) - f(x)}{h} \right| |h| \to |f'(x)| \cdot 0 = 0 $$ as $h \to 0.$ This implies that $f(x+h) \to f(x)$ as $h \to 0.$

Rewritten using $\epsilon$-$\delta$

Fix $x \in \mathbb R.$

Let $\delta_0>0$ be such that $$\left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| < 1$$ whenever $|h| < \delta_0.$ Such $\delta_0$ exists by definition of $f'(x)$ as a limit of $(f(x+h)-f(x))/h.$

Given $\epsilon>0,$ let $\delta_1 = \epsilon/(1+|f'(x)|)$ and let $\delta = \min(\delta_0, \delta_1).$ Then $$ \left| f(x+h) - f(x) \right| = \left| \frac{f(x+h) - f(x)}{h} h \right| = \left| \frac{f(x+h) - f(x)}{h} \right| |h| = \left| \frac{f(x+h) - f(x)}{h} - f'(x) + f'(x) \right| |h| \\ = \{ \text{ triangle inequality } \} \\ \leq \left( \left| \frac{f(x+h) - f(x)}{h} - f'(x) \right| + \left| f'(x) \right| \right) |h| \\ < \left( 1 + \left| f'(x) \right| \right) \frac{\epsilon}{1 + |f'(x)|} = \epsilon $$ whenever $|h| < \delta.$ Thus $f(x+h) \to f(x)$ as $h \to 0,$ i.e. $f$ is continuous in $x.$ Since $x$ was taken arbitrary, $f$ is continuous on all of $\mathbb R.$