draw a graph with a removable discontinuity.

Explanation:

Step1: Recall removable - discontinuity concept

A removable discontinuity occurs when the limit of a function exists at a point, but the function is not defined or has a different value at that point. A common example is a rational function of the form $f(x)=\frac{(x - a)(x - b)}{x - a}$ for some $a$ and $b$. Simplifying, $f(x)=x - b$ for $x\neq a$, and there is a hole at $x = a$.

Step2: Sketch the graph

1. Consider the function $y=\frac{x^{2}-1}{x - 1}$. Simplifying, $y=x + 1$ for $x\neq1$.
2. First, draw the line $y=x + 1$. The slope of the line is 1 and the $y$-intercept is 1.
3. Then, make a hole at the point $(1,2)$ (since when $x = 1$, $y=x + 1=2$ in the simplified form, but the original function $\frac{x^{2}-1}{x - 1}$ is not defined at $x = 1$).

Answer:

Sketch a line $y=x + 1$ with a hole at the point $(1,2)$.