If lim x → ∞((e/(1-e))(1/e-x/(1+x)))ˣ=α, then the value of (log ₑ α)/(1+log ₑ α) equals :

Solution

Maths sir · 4 steps

The base tends to 1 and the power to infinity, so use eˡⁱᵐˣ⁽ᵇᵃˢᵉ⁻¹⁾ for this 1^∞ form.

1. Spot the form. As x → ∞, x/(1+x) → 1, so the bracket tends to e/(1-e) × (1-e)/e = 1. The power x goes to infinity, so this is the 1^∞ form.
2. Rewrite the base. Write x/(1+x) = 1 - 1/(1+x). Then the base becomes e/(1-e)((1-e)/e + 1/(1+x)) = 1 + e/(1-e)(1+x).

   α = e^(lim(x → ∞) x · e/(1-e)(1+x)) = e^(e/(1-e))

3. Take the log. logₑ α = e/(1-e), so 1 + logₑ α = (1-e+e)/(1-e) = 1/(1-e).
4. So. (logₑ α)/(1+logₑ α) = e/(1-e) × (1-e) = e, which is option (c).

The idea to keep. Whenever the base tends to 1 and the power to infinity, the limit is eˡⁱᵐ⁽ᵖᵒʷᵉʳ⁾⁽ᵇᵃˢᵉ⁻¹⁾. Write the base as 1 + small term first.

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