Let \(f(x) = \begin{cases}\displaystyle\frac{|x - 4|}{2(x - 4)}, & x \ne 4\\ 0, & x = 4\end{cases}\). Consider the following statements:
(A) \(\displaystyle\lim_{x \to 4^-} f(x) = -\frac{1}{2}\)
(B) \(\displaystyle\lim_{x \to 4^+} f(x) = \frac{1}{2}\)
(C) f is continuous at \(x = 4\).
(D) f is continuous at \(x = 5\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
Removing the modulus:
\( \displaystyle x \gt 4: \ f(x) = \frac{x - 4}{2(x - 4)} = \frac{1}{2} \)
\( \displaystyle x \lt 4: \ f(x) = \frac{-(x - 4)}{2(x - 4)} = -\frac{1}{2} \)
So the left-hand limit at 4 is −1/2, the right-hand limit is 1/2, and f is the constant 1/2 for all x > 4.
Checking the statements
(A) \(\lim_{x \to 4^-} f(x) = -\frac{1}{2}\) — Correct. For x < 4, f(x) = −1/2.
(B) \(\lim_{x \to 4^+} f(x) = \frac{1}{2}\) — Correct. For x > 4, f(x) = 1/2.
(C) f is continuous at \(x = 4\). — Incorrect. The one-sided limits −1/2 and 1/2 are unequal.
(D) f is continuous at \(x = 5\). — Correct. Near x = 5, f is the constant 1/2.