Let the function $f : \mathbb{R} \to \mathbb{R}$, $\exists \lim \limits_{x \to c}f(x)$ for all $c\in\mathbb{R}$. Define $g : \mathbb{R} \to \mathbb{R}$ by $g(x) = \lim \limits_{t \to x}f(t)$. Show $g$ is continuous on $\mathbb{R}$.
I tried the different proof comparing the method solution suggested. From here my proof begins.
$(pf)$
Fixed $\forall c \in \mathbb{R}$, Say $\lim \limits_{x \to c}f(x) = g(c)$.
By definition,
For all $\epsilon >0$, $\exists \delta >0$ $s.t.$ $\forall x(0 < \vert x-c \vert <\delta \Rightarrow \vert f(x) – g(c) \vert < \epsilon )$ ($*$).
Considering the $ I = \{x \vert \vert x-c \vert <\delta \}$
$\vert g(x)-g(c) \vert = \vert \lim \limits_{t \to x}f(t) – \lim \limits_{t \to x}g(c) \vert = \vert \lim \limits_{t \to x}(f(t)-g(c)) \vert <\epsilon$ , $\forall x \in I$ by the $(*)$
Hence $g$ is continuous at $\forall c\in \mathbb{R}$ (In other words, $g$ is continuous on $\mathbb{R}$)
Is my proof right? If not, Please let me know which point I made mistakes.
Regards.
Best Answer
Fix $c\in \mathbb{R}$, and define the punctured neighborhood $V_{c}=(c-\delta_c,c+\delta_c)\setminus\{c\}$
$$\forall \epsilon>0, \exists \delta_c>0, \forall t\in V_{c} , \Rightarrow |f(t)-g(c)|<\frac{\epsilon}2$$
Next, $\forall x\in V_{c}$, since $\lim_{t\to x}f(t)=g(x)$, define the punctured neighborhood $V_{x}=(x-\delta_x,x+\delta_x)\setminus\{x\}$
$$\exists \delta_x>0, \forall t\in V_{x}, |f(t)-g(x)|<\frac{\epsilon}{2}$$
Finally, since $V_{c}\cap V_{x}\neq\emptyset$, take $t_1\in V_{c}\cap V_{x}$
$$\begin{align} |g(x)-g(c)|&=|g(x)-f(t_1)+f(t_1)-g(c)|\\ \\ &\le|g(x)-f(t_1)|+|f(t_1)-g(c)|<\epsilon \end{align}$$