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 $x=a$.
alternate proof of differentiable $\Rightarrow$ continuous
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calculus
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0Isn't this the standard proof? – 2012-12-06
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0Oh I see, I guess I learnt a different standard proof :) – 2012-12-06
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2Because 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 $\lim_{h\to 0}hg_a(h)=0$. – 2012-12-06
2 Answers
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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.