Let f(x)= (1+a x)^(1 / x) , x<0 ; 1+b, x=0 ; ((x+4)^(1 / 2)-2)/((x+c)^(1 / 3)-2), x>0 be continuous at x=0. Then eᵃ b c is equal to:

Solution

Maths sir · 5 steps

The right-hand piece must be a 0/0 form at 0, and that forces c = 8 before anything else.

1. Left-hand limit. As x → 0⁻, (1+ax)^(1/x) is the standard 1^∞ form, so the limit is eᵃ.
2. Fix c first. As x → 0⁺ the numerator √(x+4)-2 → 0. If c^(1/3) ≠ 2, the right-hand limit is 0, and then eᵃ = 0, which is impossible. So c^(1/3) = 2, that is c = 8.
3. Right-hand limit. Near 0, √(x+4)-2 ≈ x/4, because the derivative of (x+4)^(1/2) at 0 is 1/2√4 = 1/4. Also (x+8)^(1/3)-2 ≈ x/12, because the derivative of (x+8)^(1/3) at 0 is 1/3· 8^(-2/3) = 1/12.

   lim(x→ 0⁺)((x+4)^(1/2)-2)/((x+8)^(1/3)-2) = (1/4)/(1/12) = 3

4. Match all three. For continuity, eᵃ = 1+b = 3. This gives eᵃ = 3 and b = 2.
5. So. eᵃ b c = 3 × 2 × 8 = 48, which is option (b).

The idea to keep. When a limit must stay finite and the numerator goes to 0, the denominator must go to 0 as well. Use that to find the hidden constant first, then evaluate the 0/0 form.

Answer from the NTA final answer key. Solution by PrepWiser.