Question

If \(f(x)=\begin{cases}x+2, & x\ne0\\1, & x=0\end{cases}\), then prove that the function is not continuous at \(x=0\).

Hint:

Compute the limit of x+2 as x approaches 0 and compare it to the defined value f(0)=1.

Answer 1

Step 1: Approach from the left:
As $x\to0^-$, $f(x)=x+2\to2$.

Step 2: Approach from the right:
As $x\to0^+$, $f(x)=x+2\to2$ as well — both one-sided limits agree at 2, so the two-sided limit exists and equals 2.

Step 3: Compare the common limit to the function's defined value:
$f(0)$ is explicitly set to $1$ by the piecewise definition, not $2$.

Final Answer:
The limit exists (=2) but does not match $f(0)=1$, so continuity fails at $x=0$. \[ \boxed{\text{discontinuous at }x=0} \]