Let f x = 1 + sin ⁡ x l sin ⁡ x , - π 6 < x < 0 e tan ⁡ 2 x tan ⁡ 3 x , 0 < x < π 6 m , x = 0 is continuous at x = 0 . Then, the values of l and m are

Step-by-step solution

We have, lim h → 0 + ⁡ f 0 - h = lim h → 0 + ⁡ 1 + sin ⁡ - h l sin ⁡ - h = lim h → 0 + ⁡ 1 + sin ⁡ h l sin ⁡ h = lim h → 0 + ⁡ 1 + sin ⁡ h 1 sin ⁡ h l = e l lim h → 0 + ⁡ f 0 + h = lim h → 0 + ⁡ e tan ⁡ 2 h tan ⁡ 3 h = e lim h → 0 + ⁡ tan ⁡ 2 h 2 h × 2 3 × 3 h tan ⁡ 3 h = e 1 × 2 3 × 1 = e 2 3 Also, f 0 = m For f x to be continuous at x = 0 , we must hav