The value of
lim _(x → 0 +) (cos ⁻¹(x - [x]²).sin ⁻¹(x - [x]²))/(x - x³), where [ x ] denotes the greatest integer ≤ x is :

Solution

Maths sir · 4 steps

For x just above 0, [x] is 0, so the bracket vanishes and you are left with a standard sin⁻¹x over x limit.

1. Remove the bracket. For 0 < x < 1, the greatest integer [x] = 0. So x - [x]² = x, and the expression becomes (cos⁻¹ x · sin⁻¹ x)/(x - x³).
2. Split the fraction. Write x - x³ = x(1 - x²) and group each factor with the part it pairs with.

   lim(x → 0⁺) (sin⁻¹ x)/x · (cos⁻¹ x)/(1 - x²) = 1 · (cos⁻¹ 0)/(1 - 0)

3. Use standard limits. The standard limit gives (sin⁻¹ x)/x → 1. The other factor is continuous at 0, so cos⁻¹ x → cos⁻¹ 0 = π/2 and 1 - x² → 1.
4. So. The limit is 1 × π/2 = π/2, which is option (c).

The idea to keep. When a limit has [x], first fix which integer it equals on the side you are approaching. Then only the factor that actually goes to 0 needs a standard limit. Just substitute into the rest.

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