Show that the function f does not have limit at $x = 0:$

Question

The function $$f(x)=\left\{\begin{matrix} -1 & x\leq 0\\ 1 & x>0 \end{matrix}\right.$$

Show that the function f does not have limit at $x = 0:$

Here clearly left limit and right limit exist but not equal but how to prove this

Answer 1

This comes from the definition of the left limit and right limit. If these two limits exist, then the function has the limit in this point. Now consider that they are not equal to each other: $$\lim_{x\to0^-}f(x)=-1\neq1=\lim_{x\to0^+}f(x)$$ to prove $\lim_{x\to0^-}f(x)=-1$ with definition, we notice that for every $\varepsilon>0$ there exist $\delta=1$ such that for $-\delta<x<0$, $$|f(x)-(-1)|=|-1-(-1)|=0<\varepsilon$$ holds for all $\varepsilon>0$.

Answer 2

If the limit exists, it is either $1$, $-1$ or some other value. The $\varepsilon-\delta$ definition of the limit says that a function has a limit $L$ at $x_0$ if for every $\varepsilon>0$ exists a value $\delta>0$ such that if $|x-x_0|\le\delta$ then $|f(x)-L|\lt \varepsilon$.

If you choose for example $\varepsilon=1/2$ it means that exists a $\delta$ such that if $|x-0|=|x|\lt\delta$ you get $|f(x)-L|\lt 1/2$. And this has to hold for every $|x|\lt \delta$. We can then choose $x_+>0$ and $x_-<0$ and the result should be true. So we have $|f(x_-)-L|\lt 1/2$ and $|f(x_+)-L|\lt 1/2$. Or we can write these as $|-1-L|\lt 1/2$ and $|1-L|\lt 1/2$. The first inequality says $-3/2\lt L\lt -1/2$, the second says $1/2\lt L\lt 3/2$. Obviously the limit cannot be negative and positive at the same time, so the limit does not exists.