If lim _(x → 0) ((tan x)/x)^(1/x²)=p , then 96 log ₑ p is equal to____________

Solution

Maths sir · 3 steps

This is a 1^∞ form, so take logs and the series of tan x settles it in one line.

1. Spot the form. As x → 0, (tan x)/x → 1 and 1/x² → ∞, so the limit is of the form 1^∞. For such limits, p = e^L where L = lim (base - 1) × power.

   logₑ p = lim(x → 0) 1/x²((tan x)/x - 1) = lim(x → 0) (tan x - x)/x³

2. Use the series. Near 0, tan x = x + x³/3 + …, so tan x - x = x³/3 + …. Dividing by x³ gives logₑ p = 1/3.
3. So. 96 logₑ p = 96 × 1/3 = 32.

The common slip: 0. Treating the base as exactly 1, so that 1^∞ = 1, gives logₑ p = 0 and an answer of 0. The base only tends to 1, and the huge power magnifies the small gap.

The idea to keep. For a 1^∞ limit, write it as eˡⁱᵐ⁽ᶠ⁻¹⁾ᵍ. Then a short series like tan x = x + x³/3 usually finishes the job.

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