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

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.

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:

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

• 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 $\epsilon > 0$
– user42761
Jan 3, 2013 at 11:38
• I mean $\frac 1 {|f'(a)| + \epsilon}$
– user42761
Jan 3, 2013 at 11:50
• @André Why you take $\delta_0=\min(\delta,{1\over{|f'(a)+\epsilon}})$ instead of $1\over{|f'(a)+\epsilon}$? Jan 3, 2013 at 13:18
• 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 $$(|f'(a)|+\epsilon)|x-a| \leq (|f'(a)|+\epsilon) \frac {\epsilon}{|f'(a)|+\epsilon} \leq \epsilon$$
– user42761
Jan 3, 2013 at 15:24
• @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. Jan 3, 2013 at 19:50

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.

• This is the standard proof. However, the OP specifically asked for an $\varepsilon-\delta$ argument. Jan 4, 2013 at 18:06
• 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. Jan 4, 2013 at 18:10

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 $\delta'=\frac{\epsilon'}{2|f'(a)|}$.