Proof of continuity implied by differentiability

Question

I need to Let $f : (a, b) → R$ be differentiable on $(a, b)$. Prove that $f$ is continuous on $(a, b)$ by showing that:

$∀x_0 ∈ (a, b)$ $∀ε > 0$ $∃δ > 0$ s.t.

$(x ∈ (a, b), |x − x_0| < δ) ⇒ |f(x) − f(x_0)| < ε$

My understanding so far:

When we say a function is differentiable at $x_0$, we mean that the limit: $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.

When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$

My attempted proof:

Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} = ‎f^{\prime} ‎(x) $$

and hence

$$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$

My issue:

My issue may be a non issue, however for the 6 marks this question should be worth my proof seems shallow. Could someone please poke some holes in my proof, or add to it?

Thank you

Answer 1

Your reasoning is correct, but you might be a little relieved to use the other equivalent definition of a derivative $$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$$

Use $f(x+h) - f(x) = h \cdot \frac{f(x+h) - f(x)}{h}$, then we have

$$\lim_{h \to 0} f(x+h) - f(x) = (\lim_{h \to 0} h )\cdot f'(x) = 0 \cdot f'(x) = 0 \implies \lim_{h\to0} f(x+h) = f(x)$$

Answer 2

It is a simple and well-written proof. Perhaps that you could eliminate a tiny ambiguity: what is $\lim_{x\to x_0}f(x)-f(x_0)$? Is it $\lim_{x\to x_0}\bigl(f(x)-f(x_0)\bigr)$? Or is it $\bigl(\lim_{x\to x_0}f(x)\bigr)-f(x_0)$? So, your proof could begin with$$\bigl(\lim_{x\to x_0}f(x)\bigr)-f(x_0)=\lim_{x\to x_0}\bigl(f(x)-f(x_0)\bigr)=\cdots$$If think that this is me being pedantic, tell me and I will delete this answer.

Answer 3

Fix $\varepsilon > 0$ and $a$ arbitrarily.

From the definition of differentiation we have $$ \left|\frac{f(x)-f(a)}{x-a}-f'(a)\right| < \varepsilon $$

for an appropriately chosen $\delta > 0$.

Multiply both sides by $|x - a|$ to get: $$ \left|f(x) - f(a) - (x - a)f'(a)\right| < |x - a| \varepsilon $$

Using the reverse triangle rule, $\left||x|-|y|\right| \le |x - y|$, we have:

$$ \left|f(x) - f(a)\right| - |x - a| \cdot \left|f'(a)\right| < |x - a| \varepsilon $$

Wbhich we rearrange to get: $$ \left|f(x) - f(a)\right| < (\left|f'(a)\right| + \varepsilon) \cdot |x - a| $$

Since $f'(a)$ and $\varepsilon$ are both fixed, you can make $|f(x) - f(a)|$ as small as you want by making $|x - a|$ smaller and smaller. Thus, the function is continuous at $a$.

To prove this formally, pick any $\hat{\varepsilon}$ (different from $\varepsilon$ fixed at the beginning and used with the differentiation definition). Pick $\hat{\delta} = \min\left(\delta, \frac{\hat{\varepsilon}}{\left|f'(a)\right| + \varepsilon}\right)$. Clearly:

$$ |x - a| < \hat{\delta} \Rightarrow \left|f(x) - f(a)\right| < \hat{\varepsilon} $$

End Proof.