Answer
The graph is enclosed below.
1) At $x=-1$:
$\lim_{x\to-1^-}f(x)=0$, $\lim_{x\to-1^+}f(x)=-1$, $\lim_{x\to-1}f(x)$ does not exist.
$f(x)$ is discontinuous, but left-continuous at $x=-1$.
The discontinuity is not removable.
2) At $x=0$:
$\lim_{x\to0^-}f(x)=-\infty$, $\lim_{x\to0^+}f(x)=\infty$, $\lim_{x\to0}f(x)$ does not exist.
$f(x)$ is discontinuous at $x=0$. Neither is it left-continuous nor right-continuous here.
The discontinuity is not removable.
3) At $x=1$:
$\lim_{x\to1}f(x)=\lim_{x\to1^+}f(x)=\lim_{x\to1^-}f(x)=1$.
$f(x)$ is discontinuous at $x=1$. It is also not left-continuous or right-continuous at $x=1$.
The discontinuity is removable.
Work Step by Step
The graph is enclosed below.
*Note: $f(x)=1/x$ for $0\lt |x|\lt1$, or another way to write, $-1\lt x\lt0$ and $0\lt x\lt 1$
1) At $x=-1$:
- $\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}0=0$
- $\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}\frac{1}{x}=\frac{1}{-1}=-1$
So $\lim_{x\to-1^+}f(x)\ne\lim_{x\to-1^-}f(x)$. $\lim_{x\to-1}f(x)$ does not exist.
- $f(-1)=0$
Since $\lim_{x\to-1}f(x)$ does not exist, $f(x)$ is discontinuous at $x=-1$.
However, since $\lim_{x\to-1^-}f(x)=f(-1)=0$, $f(x)$ is left-continuous at $x=-1$.
This discontinuity is not removable because $\lim_{x\to-1}f(x)$ does not exist. We cannot remove the discontinuity by making $f(x)$ include any values so that $\lim_{x\to-1}f(x)=f(-1)$.
2) At $x=0$:
- $\lim_{x\to0^-}f(x)=\lim_{x\to0^-}\frac{1}{x}=-\infty$
- $\lim_{x\to0^+}f(x)=\lim_{x\to0^+}\frac{1}{x}=\infty$
So none of the limits exist at $x=0$.
- $f(0)$ is undefined.
Because none of the limits or $f(x)$ exist at $x=0$, $f(x)$ is discontinuous at $x=0$. It is not left-continuous or right-continuous here as well.
This discontinuity is not removable because none of the limits or $f(x)$ exist at $x=0$. We cannot remove the discontinuity by making $f(x)$ include any values so that $\lim_{x\to0}f(x)=f(0)$ as they do not exist.
3) At $x=1$:
- $\lim_{x\to1^-}f(x)=\lim_{x\to1^-}\frac{1}{x}=\frac{1}{x}=1$
- $\lim_{x\to1^+}f(x)=\lim_{x\to1^+}1=1$
So $\lim_{x\to1}f(x)=\lim_{x\to1^+}f(x)=\lim_{x\to1^-}f(x)=1$.
- $f(1)=0$
Since $\lim_{x\to1}f(x)\ne f(1)$, $f(x)$ is discontinuous at $x=1$. It is also not left-continuous or right-continuous at $x=1$.
This discontinuity is removable because only $\lim_{x\to1}f(x)\ne f(1)$. We can remove the discontinuity by making $f(x)$ include the value $f(1)=1$ so that $\lim_{x\to1}f(x)=f(1)$, thus removing the discontinuity.
