Q. 75 - Calculus

Question

Describe the intervals on which each function f  is continuous. At each point where f fails to be

continuous, use limits to determine the type of discontinuity

and any left- or right-continuity.

Step-by-Step Solution

Answer

The interval on which the function is discontinuous 

Step 1. Given information

The given information is $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2-1}{x-1}, if x<1\\ 0, if x=1\\ 3x-1, if x>1\end{array}\right.$

Step 2. Calculate the left and right limits

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{-}}{\lim }\left(\frac{x^2-1}{x-1}\right) \\ =\left(\frac{\left(1^2-1\right)}{1-1}\right) \\ =0 \end{gathered}$$

$$\begin{gathered} \underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1}{\lim }\left(0\right) \\ =0 \end{gathered}$$

$$\begin{gathered} \underset{x\to 1^{+}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(3x-1\right) \\ =\left(3\left(1\right)-1\right) \\ =2 \end{gathered}$$

since the left and right  limits both exists but not equal

hence it is discontinuous 

Step 3. The solution

Hence the function is discontinuous