Recall that a function is continuous at a point if the limit as the variable approaches a value is the same as the value of the function at that point.
Now, notice that, using the definition of the function:
[tex]\begin{gathered} \lim_{x\to1^+}f(x)=\sqrt{1}+2=3, \\ \lim_{x\to1^-}f(x)=3, \end{gathered}[/tex]
therefore:
[tex]\lim_{x\to1}f(x)=3.[/tex]
Given that the limit and the value of the function at x=1 are equal, the function is continuous at x=1.
Answer: It is continuous at x=1.
