Htykuut

Let $f: mathbb{R} to mathbb{R}$, where
$$f(x) =begin{cases}
x, & x neq 0 \
1, & x=0 \
end{cases}$$

My textbook says that $forall x_0 in mathbb{R} limlimits_{x to x_0} f(x) =x_0$. How did they reach this conclusion?

Note that $f(x)=x$ everywhere except at the single point $x=0$.

Changing the value of a continuous function on $mathbb R$ at a single point does not affect the limiting value anywhere, even at that point. That’s because when you consider $xto x_0$, you never have $x=x_0$. So here, the limit just means
$$limlimits_{xto x_0} x$$ which is just $x_0$.

Knowing the value of $f(x_0)$ tells you absolutely nothing about $limlimits_{xto x_0} f(x)$ without more information.

When talking about limits say at $x_0$, you are not interested what is going on in $x_0$ but rather around it. If you draw a graph you will see that $$limlimits_{x to 0} f(x) =0$$ for $0$ and for $x_0ne 0$ we get $$limlimits_{x to x_0} f(x) =limlimits_{x to x_0} x =x_0$$
all together: $$limlimits_{x to x_0} f(x) =x_0$$

If $x_0ne 0,$ we can find $eta>0(pmfrac{x_0}{2})$ such that
$$(forall xin]x_0-eta,x_0+eta[) ;; f(x)=x$$
thus
$$lim_{xto x_0}f(x)=lim_{xto x_0}x=x_0$$

if $x_0=0$ then

$$(forall xin(-1,0)cup(0,1));; f(x)=x$$ thus

$$lim_{xto 0,xne 0}f(x)=0=x_0$$