If f x = tan ⁡ π 4 + x 1 x , x ≠ 0 at = k , x = 0 is continuous x = 0 Then k = …

Step-by-step solution

We have, f x = tan ⁡ π 4 + x 1 x , x ≠ 0 = k , x = 0 is continuous at x = 0 ∴ l i m x → 0 f x = f 0 ⇒ l i m x → 0 tan ⁡ π 4 + x 1 x = k ⇒ l i m e x → 0 tan ⁡ π 4 + x - 1 . 1 x = k ⇒ l i m e x → 0 1 + t a n x 1 - t a n x - 1 . 1 x = k ⇒ l i m e x → 0 2 t a n x x 1 - t a n x = k ⇒ l i m e 2 x → 0 t a n x x . l i m x → 0 1 1 - t a n x = k ⇒ e 2 × 1 × 1 = k ⇒ k = e 2