Question

If the function f(x) = `[tan(π/4 + x)]^(1/x)` for x ≠ 0 is = K for x = 0 continuous at x = 0, then K = ?

Options

• e

• e^–1

• e²

• e^–2

Answer 1

e² 

Explanation:

Given, f(x) = `[tan(π/4 + x)]^(1/x)` = K

As, f(x) is continuous at x = 0,

∴ f(0) = `lim_(x rightarrow 0)` f(x)

= `lim_(x rightarrow 0)[tan(π/4 + x)]^(1/x)`

So, K = `lim_(x rightarrow 0)[(1 + tan x)/(1 - tan x)]^(1/x)` ...[1^∞ form]

= `e^(lim_(x rightarrow 0))[(1 + tan x)/(1 - tan x) - 1]1/x`

= `e^(lim_(x rightarrow 0))((2tanx)/(1 - tanx))1/x` ...`[∵ lim_(x rightarrow 0) tanx/x = 1]`

Hence K = `e^(2.1(1/(1 - 0))` = e²