Question

If f(x) = `[tan (pi/4 + x)]^(1/x)`, x ≠ 0 at

= k, x = 0 is continuous x = 0. Then k = ______.

विकल्प

• e²

• 1

• e

• e^-2

Answer 1

If f(x) = `[tan (pi/4 + x)]^(1/x)`, x ≠ 0 at

= k, x = 0 is continuous x = 0. Then k = e².

Explanation:

We have,

f(x) = `[tan (pi/4 + x)]^(1/x)`, x ≠ 0

= k, x = 0 is continuous x = 0.

`therefore lim_(x->0)`f(x) = f(0)

`=> lim_(x->0) {tan (pi/4 + x)}^(1/x)` = k

`=> "e"^(lim_(x-> 0)) [tan (pi/4 + x) - 1] * 1/x` = k

`=> "e"^(lim_(x-> 0)) ((1 + tan x)/(1 - tan x) - 1) * 1/x` = k

`=> "e"^(lim_(x-> 0)) (2 tan x)/(x (1 - tan x))` = k

`=> "e"^(lim_(2x-> 0)) (tan x)/x * lim_(x-> 0) 1/(1 - tan x)` = k

`= "e"^(2 xx 1 xx 1)` = k

⇒ k = e²