lim x → 0 e 2 sin x - 2 sin x - 1 x 2

Step-by-step solution

Given, lim x → 0 e 2 sin x - 2 sin x - 1 x 2 Now, using the expansion of e x = 1 + x 1 ! + x 2 2 ! + . . . . . . . . ∞ we get, = lim x → 0 1 + 2 sin x 1 ! + 2 sin x 2 2 ! + . . . . . ∞ - 2 sin x - 1 x 2 = lim x → 0 2 sin x 2 2 ! + 2 sin x 3 3 ! . . . . . ∞ x 2 = lim x → 0 2 sin 2 x + 2 sin x 3 3 ! . . . . . ∞ x 2 = 2 as lim x → 0 sin 2 x x 2 = 1 and other terms will become zero