PART A
We first start by computing;
[tex]\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}[/tex]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
[tex]\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}[/tex]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.
