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About derivative implies continuity: Think about this: if f is derivative at $x \in R$, then for $h \neq 0$, we have $\lim_{h \rightarrow 0}(f(x+h)-f(x))=0$.

My question is how to proof the formula $\lim_{h \rightarrow 0}(f(x+h)-f(x))=0 \implies \lim_{h \rightarrow 0}^{}f(x+h)=f(x)$ i.e. f is continuous at $x \in R$?

Fu Yu Cheng
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3 Answers3

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The equation $$\lim_{h\to0} \big( f(x+h)-f(x) \big) = 0 \tag{1}$$ says that the distance between $f(x+h)-f(x)$ and $0$ is small as $h$ is sufficiently close to $0$. More formally, given an arbitrary $\varepsilon>0$, $$\big|\big( f(x+h)-f(x) \big)-0 \big| < \varepsilon \quad \textrm{as } h\in(-\delta,\delta) \tag{2}$$ for some $\delta>0$. Here, you should be interpret $|x-y|$ as the distance between two real numbers $x$ and $y$. That is, $(2)$ says that the distance between $f(x+h)-f(x)$ and $0$ is less than $\varepsilon$, for every $\varepsilon$ as you want, whenever $h$ is in some open neighborhood of $0$.

Also, the equation $$\lim_{h\to0} f(x+h) = f(x) \tag{3}$$ says that, for every $\varepsilon >0$, the distance between $f(x+h)$ and $f(x)$ is less than $\varepsilon$ whenever $h$ is in some neighborhood of $0$, that is, $$|f(x+h)-f(x)| < \varepsilon \textrm{ whenever } h\in(-\delta,\delta) \tag{4}$$ for some $\delta > 0$. Of course, there are no difference between $(2)$ and $(4)$ since $$|f(x+h)-f(x)| = \big|\big( f(x+h)-f(x) \big)-0 \big|.$$ So, in fact, $(1)$ and $(3)$ they say exactly the same, in other words, they are equivalent.

azif00
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  • The formula $h∈(−δ,δ)$, I think it should rewrite to $h∈(−δ,0) \cup (0,δ)$, i.e. $0 \notin (−δ,δ)$. – Fu Yu Cheng Jan 03 '20 at 06:02
  • That is in the standard definition of a limit. In the definition of continuity doesn't matter, since $f$ is defined at $x$, which is the same as saying that the function $g$ given by $g(h) = f(x+h)$ is defined at $h=0$. – azif00 Jan 03 '20 at 06:24
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We have $\lim_{h\to 0} f(x+h) - f(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \cdot h = \lim_{h\to 0} f'(x) \cdot h = f'(x) \cdot 0 = 0$

So for every $\varepsilon > 0$, there exist $\delta > 0$ such that $|h|<\delta \implies $

$|f(x+h)-f(x) - 0| <\varepsilon$

i.e. $|f(x+h)-f(x)|<\varepsilon$

And well, this is the definition of $\lim_{h \to 0} f(x+h) = f(x)$

ZAF
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  • OP is asking how to prove that $$\lim_{h\to0} \big( f(x+h)-f(x) \big) = 0 \textrm{ implies } \lim_{h\to0} f(x+h) = f(x).$$ – azif00 Jan 03 '20 at 03:23
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For an alternate proof that differentiability implies continuity, we can use the equivalence of sequential continuity with continuity in metric spaces (in fact this is true for any topological space in which finite sets are closed). Let $x_n$ be a sequence of elements of $\mathbb R$ that converge to $x$ in $\mathbb R$. Then $$ \lim_{n\to\infty}\frac{f(x_n)-f(x)}{x_n-x} = 0, $$ so in particular $$ \lim_{n\to\infty} f(x_n)-f(x) = 0, \text{ or } \lim_{n\to\infty} f(x_n) = f(x). $$ This follows from $\liminf_{n\to\infty}|x_n-x|<1$.

Math1000
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