$$\lim_{h\to0}f(a+h)-f(a)=0\leftrightarrow \lim_{x\to a}f(x)=f(a)$$
Proof.
$\Rightarrow:$
Assume
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|\color{blue}{h}|<\delta\rightarrow |f(a+\color{blue}{h})-f(a)|<\varepsilon$$
Show
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|x-a|<\delta\rightarrow |f(x)-f(a)|<\varepsilon$$
Let $h:=x-a$ that by assumption we have
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|\color{blue}{x-a}|<\delta\rightarrow |f(a+\color{blue}{x-a})-f(a)|<\varepsilon$$
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|\color{blue}{x-a}|<\delta\rightarrow |f(x)-f(a)|<\varepsilon$$
$\Leftarrow:$
Assume
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|\color{blue}{x-a}|<\delta\rightarrow |f(x)-f(a)|<\varepsilon$$
Show
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|h|<\delta\rightarrow |f(a+h)-f(a)|<\varepsilon$$
Let $x-a:=h$ by assumption we have
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|\color{blue}{x-a}|<\delta\rightarrow |f(a+\color{blue}{x-a})-f(a)|<\varepsilon$$
$$\forall \varepsilon>0,\exists\delta>0,s.t.0<|\color{blue}{h}|<\delta\rightarrow |f(a+\color{blue}{h})-f(a)|<\varepsilon\tag*{$\square$}$$
I saw some notes state that $\lim_{h\to0}f(a+h)-f(a)=0$ implies continuous, but i
don't know this alternative definition before, so i'm tring to prove this.
the proof seems like just plug in some variables to the limit-definition
Is this correct, thanks for your help.
Best Answer
We have that the two statements are equivalent, indeed
$$\lim_{h\to0}f(a+h)-f(a)=0 \iff\lim_{h\to0}f(a+h)=f(a)$$
and by $x=a+h \to a$
$$\lim_{h\to0}f(a+h)=f(a) \iff \lim_{x\to a}f(x)=f(a)$$
that is the definition of continuous function.