Is there a function which approaches the same value from both sides but is defined differently at that point? e.g. a function, f(x) where f(0) approaches 0 from positive x and 0 from negative x, but when you actually compute f(0) it is defined as a different value like 1? And I don't mean some function where you define it differently for x > 0, x < 0 and x = 0.
Additionally a function where the limit at an x approaches a different number from positive x than from negative x, but is actually defined at that point. When the limit approaches a different value from two different directions usually we say it is therefore undefined at that point, but is this always the case?
 A: Let $f:\mathbb R\setminus \{0\}\to \mathbb R$ be a function such that $f(x)=0$.
This function meets all your requirements. First, $f(x)\rightarrow 0$ when $x \rightarrow 0^-$ and $f(x)\rightarrow 0$ when $x \rightarrow 0^+$ . Second, when you compute $f(x)$ at $x=0$ (which you are not supposed to do) i.e., $f(0)$, it is undefined because I defined the function in such a domain that it is undefined at $x=0$. Third, it is not defined differently for $x>0$ and $x<0$.
Is it defined differently for $x=0$? Absurd question! You yourself said $f$ should not be defined at $x=0$ but it takes values close to zero when $x$ approaches zero. So I need to define the function in $\{0-\epsilon, 0+\epsilon\} \setminus \{0\}$, where $\epsilon >0$ is any arbitrary real number
A: What do you mean by "is not defined"? We can always define a function on a subset. If you mean as in forbidden because dividing by 0, then the $f(x) = x+4 = (x^2+4x)/x$ is a good example for a discontinuity.
But maybe what you are looking for is $f(x) = \sin(1/|x|)$ , which does not behave well at 0.
A: I just thought of this one: $y=0^{|x|}$. From this you can then have $y=0^{|x|} + x^2$ and so on.
