Q. 78 - 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.

$f\left(x\right)=\left\{\begin{array}{ll}{2}^x-1,& if x\le 0\\ \frac{4^x-1}{2^x-1}& if x>0\end{array}\right.$

Step-by-Step Solution

Answer

The function is continuous on R.

Step 1. Given information.

Consider the given question,

$f\left(x\right)=\left\{\begin{array}{ll}{2}^x-1,& if x\le 0\\ \frac{4^x-1}{2^x-1}& if x>0\end{array}\right.$

Step 2. Calculate the left and right limits.

We need to calculate the left and right limits,

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

As the left and right  limits both exists but are equal.

Hence, the function is continuous on R.