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: Find the limit as \(x\to0\):
For \(x\ne0\), \(f(x)=x+2\), so \(\displaystyle\lim_{x\to0}f(x)=0+2=2\).

Step 2: Find the function's actual value at \(x=0\):
By definition, \(f(0)=1\).

Step 3: Compare:
Continuity at \(x=0\) requires \(\displaystyle\lim_{x\to0}f(x)=f(0)\). Here \(2\ne1\).

Final Answer:
Since the limit and the function value disagree, \(f\) is not continuous at \(x=0\). \[ \boxed{\lim_{x\to0}f(x)=2\ne f(0)=1} \]