# alternate proof of differentiable $\Rightarrow$ continuous

I'm wondering if the following reasoning is correct: Suppose that $f(x)$ is differentiable at $a$. Then $\lim_{h\rightarrow 0}(f(a+h) - f(h))/h$ must exist. Since $\lim_{h\rightarrow 0} h =0$, it follows that $$\lim_{h\rightarrow 0} (f(a+h) - f(a)) = 0,$$ otherwise the derivative at $0$ does not exist. This implies $$\lim_{h\rightarrow 0} f(a+h) = \lim_{h\rightarrow 0} f(a),$$ which shows that $f(x)$ is continuous at $x=a$.

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Isn't this the standard proof? –  N. S. Dec 6 '12 at 5:24
Oh I see, I guess I learnt a different standard proof :) –  somebody Dec 6 '12 at 5:29
Because reasoning about limits when ratios are involved can be treacherous, I prefer a more standard presentation. For $h\ne 0$, let $\frac{f(a+h)-f(a)}{h}=g_a(h)$. So $f(a+h)=f(a)+hg_a(h)$. By differentiability, $\lim_{h\to 0}g_a(h)$ exists. Thus $\lim_{h\to 0}hg_a(h)=0$. –  André Nicolas Dec 6 '12 at 5:39

Given

$$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$

exists. Now, note that

$$(f(x+h)-f(x))= \left(\frac{f(x+h)-f(x)}{h}\right)h$$

$$\implies \lim_{h\to 0}(f(x+h)-f(x))= \lim_{h \to 0}\left(\frac{f(x+h)-f(x)}{h}\right)\lim_{h \to 0} h = f'(x).0 = 0,$$

and the result follows.

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That's entirely correct. In the contrapositive: if $f$ is discontinuous at $a$, then $$\lim f(a+h)-f(a)\neq 0.$$

This means there are some $\varepsilon,\delta>0$ so that when $x$ is $\delta$-close to $a$, $|f(a+h)-f(a)|>\varepsilon$, so

$$\frac {|f(a+h)-f(a)|}{h}>\frac\varepsilon h,$$

which becomes arbitrarily large as $h$ gets small.

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