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How do you prove that differentiability implies continuity with $\epsilon$-$\delta$ definition?

I know that's a very common theorem in calculus but when I try to prove it with $\epsilon$-$\delta$ definition of continuity, I found that it is not so obvious.

Attempt: Let $f:\mathbb{R}\to\mathbb{R}$ be a function differentiable at point $$a \implies \forall \epsilon>0, \exists \delta>0 \text{ s.t. } \left|\frac{f(x)-f(a)}{x-a}-f'(a)\right|<\epsilon$$ for any $|x-a|<\delta$. So what we want to show is that for all $\epsilon>0$, we can find an $\delta>0$ such that $|f(x)-f(a)|<\epsilon$ for any $|x-a|< \delta$. First of all, we can apply the triangula inequality $$|f(x)-f(a)|\le |f(x)-f(a)-f'(a)(x-a)|+|f'(a)(x-a)|<\epsilon+|f'(a)(x-a)|,$$ but I found that $|f'(a)(x-a)|$ could be very large even $\epsilon$ can be any real number.

Xander Henderson
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Mathematics
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3 Answers3

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

From the definition of differentiation we have $$ \left|\frac{f(x)-f(a)}{x-a}-f'(a)\right| < \varepsilon\tag{1}$$

for an appropriately chosen $\delta > 0$.

Multiply both sides by $|x - a|$ to get: $$ \left|f(x) - f(a) - (x - a)f'(a)\right| < |x - a| \varepsilon\tag{2} $$

Using $\left||x|-|y|\right| \le |x - y|$ we have:

$$ \left|f(x) - f(a)\right| - |x - a| \cdot \left|f'(a)\right| < |x - a| \varepsilon\tag{3} $$

Rearrange to get: $$ \left|f(x) - f(a)\right| < (\left|f'(a)\right| + \varepsilon) \cdot |x - a|\tag{4} $$

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:

$$ |x - a| < \hat{\delta} \Rightarrow \left|f(x) - f(a)\right| < \hat{\varepsilon}\tag{5} $$

lineage
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Ayman Hourieh
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    Foor choosing $\delta > 0$ you can think about $\delta_0 := min(\delta,\epsilon \cdot \frac 1 {|f'(x)|+\epsilon})$ where $\delta > 0$ is such that $$\left | \frac { f(x)-f(a) } {x-a} - f'(a) \right | < \epsilon $$ for $|x-a| < \delta$ given $\epsilon > 0$ –  Jan 03 '13 at 11:38
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    I mean $\frac 1 {|f'(a)| + \epsilon} $ –  Jan 03 '13 at 11:50
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    @André Why you take $\delta_0=\min(\delta,{1\over{|f'(a)+\epsilon}})$ instead of $1\over{|f'(a)+\epsilon}$? – Mathematics Jan 03 '13 at 13:18
  • @Mathematics - I added the final $\varepsilon-\delta$ argument to my answer above. The inequality I derive holds when $|x - a| < \delta$. Therefore, we should ensure that $x$ continues to satisfy this when making the continuity $\varepsilon-\delta$ argument. – Ayman Hourieh Jan 03 '13 at 14:20
  • @AymanHourieh My question above the comment is why you don't just take $\delta_0={1\over {|f'(a)|+\epsilon$}}$ but consider the $\min$ ? – Mathematics Jan 03 '13 at 14:34
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    You first coose $\epsilon > 0$ arbitrary. Then ther is a $\delta > 0$ such that $|(f(x)-f(a))/(x-a)|< \epsilon$ for $|x-a| \leq \delta$. By letting $|x-a| \leq \delta_0$ you ensure that $|x-a| \leq \delta$ and $|x-a| \leq \frac{\epsilon}{|f'(a)|+\epsilon}$ such that $$ (|f'(a)|+\epsilon)|x-a| \leq (|f'(a)|+\epsilon) \frac {\epsilon}{|f'(a)|+\epsilon} \leq \epsilon$$ –  Jan 03 '13 at 15:24
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    @Mathematics We take the min because we don't know which is smaller, $\delta$ or $\frac{\hat{\varepsilon}}{\left|f'(a)\right| + \varepsilon}$. The two values are independent of each other, and we want $x$ to satisfy both conditions. – Ayman Hourieh Jan 03 '13 at 19:50
  • Why do I need to fix $\varepsilon$, that is why can't I take $\delta = \varepsilon /(f^\prime (a)+ \varepsilon)$ where $\delta \leq \delta^{ \prime}$ used in differentiability. – Kareem Taha Oct 19 '20 at 09:21
  • My derivation uses this, $\left|f(x) - f(a)\right| < (\left|f'(a)\right| + \varepsilon) \cdot |x - a|< (\left|f'(a)\right| + \varepsilon) \cdot \delta$. Holds for any $\delta\leq \delta^{\prime}$. – Kareem Taha Oct 19 '20 at 09:30
  • In the third equation from top, where you used modulus lower bound, how did you you resolve the outer modulus on the LHS ? i.e. shouldn't the LHS be $\left |\left|f(x) - f(a)\right| - |x - a| \cdot \left|f'(a)\right| \right| < |x - a| \varepsilon$ – lineage Dec 06 '23 at 14:31
  • @lineage $$|u|-|v|=|u-v+v|-|v|\leq|u-v|+|v|-|v|=|u-v|$$ $$u=f(x)-f(a),\quad v=(x-a)f'(a)$$ – mr_e_man Dec 06 '23 at 15:42
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You want to show that $d(f(x),f(t))<\epsilon$ when $d(x,t)<\delta$.

$\displaystyle\lim_{x\to t}f(x)-f(t)= \lim_{x\to t}\frac{f(x)-f(t)}{x-t}(x-t)=f'(t)\cdot0=0$ which is what we wanted to show.

user10444
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    This is the standard proof. However, the OP specifically asked for an $\varepsilon-\delta$ argument. – Ayman Hourieh Jan 04 '13 at 18:06
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    This is a nice, simple, direct proof; one of my favorites. It is the same proof that Rudin uses in his "Principles of Mathematical Analysis." I think it is easy to rewrite this as an epsilon-delta proof. – Chris Leary Jan 04 '13 at 18:10
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You don't need to add much to your own proof to finish it off. You've shown that, given $\epsilon>0$, you can find $\delta>0$ such that $|f(x)-f(a)|<\epsilon+|f'(a)||x-a|$ whenever $|x-a|<\delta.$ This means $|f(x)-f(a)|<\epsilon+|f'(a)|\delta$ whenever $|x-a|<\delta$.

Now, given any $\epsilon'>0$. We want to show that there exists $\delta'>0$ such that $|f(x)-f(a)|<\epsilon'$ whenever $|x-a|<\delta'.$ If $f'(a)=0$ then, letting $\epsilon=\epsilon',$ we know there is a $\delta$ such that $|f(x)-f(a)|<\epsilon'$, so we take $\delta'=\delta$. Otherwise, let $\epsilon=\frac{\epsilon'}{2}$ so that there exists $\delta$ such that $|f(x)-f(a)|<\frac{\epsilon'}{2}+|f'(a)|\delta$. If $\delta\le\frac{\epsilon'}{2|f'(a)|}$, then the right side is less than or equal to $\epsilon'$, so we let $\delta'=\delta$. Otherwise we let $\delta'=\frac{\epsilon'}{2|f'(a)|}$.

Will Orrick
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