The value of
lim _(x → 0) ( x/(⁸√(1 - sin x) - ⁸√(1 + sin x)) ) is equal to :

Solution

Maths sir · 4 steps

Near zero, an eighth root of 1 plus something small is about 1 plus one eighth of that something.

1. Spot the form. At x = 0 the top is 0 and the bottom is 1 - 1 = 0. This is a 0/0 form, so the denominator has to be reshaped.
2. Binomial approximation. For small t, (1+t)ⁿ ≈ 1 + nt. With n = 1/8 this gives (1-sin x)^(1/8) ≈ 1 - (sin x)/8 and (1+sin x)^(1/8) ≈ 1 + (sin x)/8.

   ⁸√(1-sin x)-⁸√(1+sin x) ≈ -(2 sin x)/8 = -(sin x)/4

3. Use the standard limit. The expression becomes x/(-(sin x)/4) = -4 × x/(sin x). Since x/(sin x) → 1 as x → 0, the limit is -4 × 1.
4. So. The limit equals -4, which is option (c).

The idea to keep. For a difference of nth roots near 1, use (1+t)^(1/n) ≈ 1 + t/n. The two 1s cancel and the small parts decide the limit. Keep the order of subtraction to get the sign right.

Answer from the NTA final answer key. Solution by PrepWiser.