Given below are two statements:
Statement I: lim(x → 0) ( (tan⁻¹ x + logₑ √((1+x)/(1-x)) - 2x)/x⁵ ) = 2/5
Statement II: lim(x → 1) ( x^(2/(1-x)) ) = 1/e²
In the light of the above statements, choose the correct answer from the options given below:

Solution

Maths sir · 4 steps

Expand both terms of Statement I as series and watch them cancel; Statement II is the classic 1^∞ form.

1. Two series. Near x = 0, tan⁻¹ x = x - x³/3 + x⁵/5 - … For the log, logₑ √((1+x)/(1-x)) = 1/2[logₑ(1+x) - logₑ(1-x)] = x + x³/3 + x⁵/5 + …

   tan⁻¹ x + logₑ √((1+x)/(1-x)) - 2x = 2x⁵/5 + (terms in x⁷ and higher)

2. Statement I. The x terms cancel against -2x, and the x³ terms cancel each other. Dividing by x⁵ leaves 2/5 as x → 0, so Statement I is true.
3. Statement II. As x → 1 the base tends to 1 and the power 2/(1-x) blows up, so this is a 1^∞ form. The limit is e^L with L = lim(x → 1) (x - 1) × 2/(1-x) = -2. So the limit is e⁻² = 1/e², and Statement II is true.
4. So. Both statements are true, which is option (c).

The idea to keep. When several series terms sit in the numerator over xⁿ, expand each one up to xⁿ and cancel. For 1^∞, use lim fᵍ = eˡⁱᵐᵍ⁽ᶠ⁻¹⁾.

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