Let f : R → R be defined as f(x) = { x³/(1 - cos 2x)²log ₑ( (1 + 2xe⁻²ˣ)/(1 - xe⁻ˣ)² ),, x ≠ 0; α ,, x = 0
If f is continuous at x = 0, then α is equal to :

Solution

Maths sir · 4 steps

Count the powers of x: the bottom behaves like 4x⁴, so the log must supply exactly 4x.

1. Size of the denominator. 1 - cos 2x = 2 sin² x, so (1-cos 2x)² = 4 sin⁴ x. Near 0 this behaves like 4x⁴, so x³/(4 sin⁴ x) blows up like 1/4x.
2. Split the log. The log of the fraction is ln(1+2xe⁻²ˣ) - 2 ln(1-xe⁻ˣ). Each piece goes to 0, so this is a 0 × ∞ form. Pair the extra 1/x with the log.

   f(x) = x⁴/(4 sin⁴ x)[ln(1+2xe⁻²ˣ)/2xe⁻²ˣ· 2e⁻²ˣ + 2·ln(1-xe⁻ˣ)/-xe⁻ˣ· e⁻ˣ]

3. Take the limit. Use ln(1+u)/u → 1 as u → 0 and x/(sin x) → 1. The bracket tends to 2 × 1 + 2 × 1 = 4, and the front factor tends to 1/4.
4. So. lim(x → 0) f(x) = 1/4 × 4 = 1. Continuity at 0 needs α to equal this limit, so α = 1, option (a).

The idea to keep. In a 0 × ∞ limit, find the order of each factor first. Replace 1-cos 2x by 2 sin² x and turn every log into the standard form ln(1+u)/u → 1.

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