lim _(x → 0) ( tan ( π/4 + x ) )^(1/x) is equal to :

Solution

Maths sir · 4 steps

The base goes to 1 and the power blows up, so this is 1^∞: use eˡⁱᵐᵍ⁽ᶠ⁻¹⁾.

1. Spot the form. As x → 0, tan(π/4 + x) → tan π/4 = 1 and 1/x → ∞. This is the 1^∞ form, so the limit equals e^L with L = lim g(x) (f(x) - 1).

   L = lim(x → 0) 1/x(tan(π/4+x) - 1)

2. Expand the tangent. tan(π/4+x) = (1+tan x)/(1-tan x). Subtracting 1 gives (2 tan x)/(1-tan x).
3. Find L. L = lim(x → 0) (tan x)/x × 2/(1-tan x) = 1 × 2/(1-0) = 2.
4. So. The limit is e^L = e², which is option (d).

The idea to keep. For any 1^∞ limit fᵍ, the answer is eˡⁱᵐᵍ⁽ᶠ⁻¹⁾. Find that exponent first, and do not stop until you have raised e to it.

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