Finding an appropriate upper bound for $\frac{x^2-4}{x-1}$ at $x=0$ and $x=2$ I am investigating the continuity of the function $\frac{x^2-4}{x-1}$ at $x=0$ and  $x=2$.
Here's a part of my proof;

$$\lvert f(x)-f(0)\rvert =\left\lvert\frac{x^2-4}{x-1}-4\right\rvert$$
  $$=\left\lvert \frac{x(x+4)}{x-1}\right\rvert$$
  $$=\left\lvert \frac{x(x+4)}{x-1}\right\rvert$$
  $$=\frac{\left\lvert x\right\rvert\left\lvert x+4\right\rvert}{\left\lvert x-1\right\rvert}$$

The challenge I have is to find an appropriate upper bound for $\frac{1}{x-1}$ at $x=0$ and $x=2$. Please, can anyone help out?


$$\lvert f(x)-f(0)\rvert =\frac{\left\lvert x\right\rvert\left\lvert x+4\right\rvert}{\left\lvert x-1\right\rvert}$$
The challenge I have is to find an appropriate upper bound for $\frac{1}{x-1}$ at $x=0$ and $x=2$. Please, can anyone help out?

You want to stay close enough to $x=0$ to ensure $\tfrac{1}{x-1}$ stays bounded.
You obviously don't want to go near $x=1$, since $\tfrac{1}{x-1}$ then becomes (arbitrary) large. Limit yourself to a distance of e.g. $\tfrac{1}{2}$ from $x=0$, then:
$$|x|<\tfrac{1}{2} \iff -\tfrac{1}{2} < x < \tfrac{1}{2}\iff -\tfrac{3}{2} < x-1 < -\tfrac{1}{2}$$
What does this tell you about $|x-1|$ and thus about $\tfrac{1}{|x-1|}$?
