Question

Is the function \(f(x)=\begin{cases}x+5,&x\le1\\x-5,&x>1\end{cases}\) continuous at \(x=1\)?

Hint:

Compare LHL, RHL and f(1); a mismatch means discontinuous.

Answer 1

Step 1: Key Formula or Approach:
For a piecewise function at the joining point, check whether the left-hand limit (LHL), right-hand limit (RHL) and the function value all agree. If any two of these disagree, the function is discontinuous there.

Step 2: Finding \(f(1)\):
Since \(1\le1\), use the first branch: \(f(1) = 1+5 = 6\).

Step 3: Finding the left-hand limit:
For \(x\) slightly less than \(1\), use \(f(x)=x+5\):
\[ \lim_{x\to1^-}f(x) = 1+5 = 6 \]

Step 4: Finding the right-hand limit:
For \(x\) slightly more than \(1\), use \(f(x)=x-5\):
\[ \lim_{x\to1^+}f(x) = 1-5 = -4 \]

Step 5: Comparing:
LHL \(=6\) but RHL \(=-4\); these are not equal, so the two-sided limit does not exist at \(x=1\).

Final Answer:
Since LHL \(\ne\) RHL, \(f(x)\) is NOT continuous at \(x=1\). \[ \boxed{\text{No, } f \text{ is discontinuous at } x=1 \text{ (LHL}=6\ne\text{RHL}=-4)} \]