Question

Calculus - determine the continuity

Answer 1

PART A

We first start by computing;

`\begin{gathered} \text{For c=0} \\ \lim _(x\to c^-)f(x)=x^2+1=0^2+1=1 \\ \lim _(x\to c^+)f(x)=\sqrt[]{x}+1=\sqrt[]{0}+1=1 \end{gathered}`

From the computing above "f " is removable discontinuous because f is not defined at zero. That is f(0) does not exist. This can also be seen in the graph plotted for all the functions used in finding the limit.

PART B

`\begin{gathered} \text{For c=1} \\ \lim _(x\to c^-)f(x)=\sqrt[]{x}^{}+1=1^{}+1=2 \\ \lim _(x\to c^+)f(x)=2x=2(1)=2 \end{gathered}`

From the computing above "f " is continuous because f is defined at one. That is f(1) exist. This can also be seen in the graph plotted for all the functions used in finding the limit.