Q165 / 240 · KCET 2026 · Mathematics
If $f(x) = \begin{cases} x^2 - 1 & \text{if } x \geq 2 \\ x + 1 & \text{if } x < 2 \end{cases}$, then $\lim\limits_{x \to 1} f(x) + \lim\limits_{x \to 2} f(x) = $
- A$3$
- Correct answerB$5$
- C$7$
- D$9$
Solution
For $x \to 1$, since $x < 2$, the function is $f(x) = x + 1$.
$\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (x + 1) = 1 + 1 = 2$
For $x \to 2$, we evaluate the left-hand and right-hand limits.
Left-hand limit at $x = 2$:
$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (x + 1) = 2 + 1 = 3$
Right-hand limit at $x = 2$:
$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (x^2 - 1) = 2^2 - 1 = 3$
Since the left-hand limit equals the right-hand limit, $\lim\limits_{x \to 2} f(x) = 3$.
The required sum is $\lim\limits_{x \to 1} f(x) + \lim\limits_{x \to 2} f(x) = 2 + 3 = 5$.
Answer: $5$
$\lim\limits_{x \to 1} f(x) = \lim\limits_{x \to 1} (x + 1) = 1 + 1 = 2$
For $x \to 2$, we evaluate the left-hand and right-hand limits.
Left-hand limit at $x = 2$:
$\lim\limits_{x \to 2^-} f(x) = \lim\limits_{x \to 2^-} (x + 1) = 2 + 1 = 3$
Right-hand limit at $x = 2$:
$\lim\limits_{x \to 2^+} f(x) = \lim\limits_{x \to 2^+} (x^2 - 1) = 2^2 - 1 = 3$
Since the left-hand limit equals the right-hand limit, $\lim\limits_{x \to 2} f(x) = 3$.
The required sum is $\lim\limits_{x \to 1} f(x) + \lim\limits_{x \to 2} f(x) = 2 + 3 = 5$.
Answer: $5$
Hence, the correct answer is option: 'B'.