$\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}}(\sin \sqrt{t}) d t}{x^{3}}$ is equal to:

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Limits

lim_{x \to 0} \frac{\int _{0}^{x^{2}} ( s i n t ) d t}{x ^{3}}

is equal to:

Answer: (B)

x \to 0

the integral in the numerator and the denominator

x^{3}

both vanish, giving an indeterminate ratio. Near

t = 0

the integrand satisfies

sin t \approx t

, so for small

x

the numerator is close to

\int_{0}^{x^{2}} t d t = \frac{2}{3} (x^{2})^{3/2} = \frac{2}{3} x^{3}

, and dividing by

x^{3}

leaves

\frac{2}{3}

. The same value follows from differentiating numerator and denominator: the numerator's derivative is

2 x sin ∣ x ∣

and the denominator's is

3 x^{2}

, and

\frac{2 x sin ∣ x ∣}{3 x ^{2}} \to \frac{2}{3}

x \to 0

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option

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