Given:
f(x) =
2x + 3, x ≤ 0
3(x + 1), x > 0
Part (i): Evaluate limx→0 f(x)
Step 1: Left Hand Limit (LHL)
LHL = limx→0⁻ f(x)
= limx→0⁻ (2x + 3)
= 3
Step 2: Right Hand Limit (RHL)
RHL = limx→0⁺ f(x)
= limx→0⁺ 3(x + 1)
= 3
Step 3: Compare LHL and RHL
Since LHL = RHL = 3,
limx→0 f(x) = 3
Part (ii): Evaluate limx→1 f(x)
For x → 1, we have x > 0,
So, f(x) = 3(x + 1)
limx→1 f(x) = 3(1 + 1)
= 6
Final Answer:
limx→0 f(x) = 3
limx→1 f(x) = 6
The area of the region \( \{(x, y): 0 \leq y \leq x^2 + 1, \, 0 \leq y \leq x + 1, \, 0 \leq x \leq 2\ \) is:}