Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $x=a$.

share|cite|improve this question
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
up vote 2 down vote accepted


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

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.