lim _(x → 0) (∫₀^(x²) ( sin √t )dt)/x³ is equal to :

Solution

Maths sir · 4 steps

An integral with x in its upper limit over a power of x means L'Hôpital, with the Leibniz rule on top.

1. Spot the form. At x = 0 the integral runs from 0 to 0, so the numerator is 0. The denominator x³ is also 0. This is the 0/0 form, so L'Hôpital's rule applies.
2. Differentiate the top. By the Leibniz rule, the derivative of ∫₀ᵍ⁽ˣ⁾ f(t) dt is f(g(x)) g'(x). With g(x) = x² this gives sin(√x²) × 2x. The expected answer takes √x² = x, so the derivative is 2x sin x. The bottom differentiates to 3x².

   lim(x → 0) (2x sin x)/3x² = 2/3 lim(x → 0) (sin x)/x = 2/3

3. A careful note. Strictly, √x² = |x|, so the ratio becomes 2/3 × sin|x|/x. This tends to 2/3 from the right and -2/3 from the left. The answer key uses the right-hand value, x → 0⁺.
4. So. The limit is 2/3, which is option (c).

The idea to keep. A variable limit of integration over a power of x almost always gives 0/0. Differentiate the integral as f(upper limit) × (upper limit)'. Do not forget the chain factor.

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