Question

Determine if r(x) is continuous at 1.26? Use limits in your explanation of why this functions is or is not continuous at 1.26

Answer 1

Given:

$R(x)\text{ = }\frac{x^3-x^2+x-1}{x^4-x^3+2x-2}$

A function f(x) is continuous at a point x=c :

$\lim_{x\to c}\text{ f\lparen x\rparen = f\lparen c\rparen}$

First, let us take the limit:

$=\lim_{x\to\:1.26}\left(\frac{x^3-x^2+x-1}{x^4-x^3+2x-2}\right)$

Plug in the values x = 1.26

$\begin{gathered} =\frac{1.26^3-1.26^2+1.26-1}{1.26^4-1.26^3+2\cdot \:1.26-2} \\ =0.64683 \end{gathered}$

Now, let us find R(1,26):

$R(1.26)\text{ = 0.6468}$

Hence, we can conclude that R(x) is continuous at x = 1.26