Science:Math Exam Resources/Courses/MATH110/December 2013/Question 01 (c)/Solution 1

False: To hunt for discontinuities, we have to check inside each piece and also at the boundary.

• For the boundary to be continuous, we must have:

  $\begin{array}{rl}\underset{x\to 1^{-}}{\lim }f(x)& =\underset{x\to 1^{+}}{\lim }f(x)\\ \underset{x\to 1^{-}}{\lim }\frac{3}{x+2}& =\underset{x\to 1^{+}}{\lim }\sqrt{x}\\ \frac{3}{1+2}& =\sqrt{1}\\ 1& =1.\end{array}$

  So the limits exists and is equal to $1$ which happens to be $f(1)=1$. So the function in continuous at the boundary.

• On the right of $x=1$, the function is continuous since $\sqrt{x}$ is only undefined for $x<0$ which is not covered by this case.

• On the left of $x=1$, the function is discontinuous at $x=-2$ since the denominator is $0$. This fall inside the region considered. So $f(x)$ is discontinuous at $x=-2$.

That means $f(x)$ is NOT continuous over all real numbers.