Q. 19 - Calculus

Question

Use a calculator or other graphing utility to graph the function $g\left(x\right)=\frac{x^2-2x+1}{x-1}$. 

(a) Show that g(x) is not defined at x = 1. How is this reflected in your calculator graph? 

(b) Use the graph to argue that even though g(1) is undefined, we have $\underset{x\to 1}{lim} g\left(x\right)=0$. 

Step-by-Step Solution

Answer

Part (a). 

Part (b). 0

Part (a) Step 1. Given information.

We have been given a function $g\left(x\right)=\frac{x^2-2x+1}{x-1}$.

We have to use a calculator or other graphing utility to graph this function. 

Show that g(x) is not defined at x = 1. How this is reflected in the calculator graph.

Part (a) Step 2. Sketch the graph.

Function is :

$g\left(x\right)=\frac{x^2-2x+1}{x-1}$

The graph of the function is :

Putting x=1 in the function, we get:

$\begin{array}{r}g\left(1\right)=\frac{1^2-2\left(1\right)+1}{1-1}\\ =\frac{1-2+1}{0}\\ =\frac{0}{0}\end{array}$

                         = Undefined

It is not immediately clear that g(x) IS not defined at x=1

Part (b) Step 1. Solving the limit expression.

Solving the limit expression, 

$\begin{array}{r}\underset{x\to 1}{lim} g\left(x\right)=\underset{x\to 1}{lim} \frac{x^2-2x+1}{x-1}\\ =\underset{x\to 1}{lim} \frac{(x-1{)}^2}{x-1}\\ =\underset{x\to 1}{lim} (x-1)\\ =1-1\\ =0\end{array}$