Let $f(x)=\int_{0}^{x}(t+\sin(1-e^{t}))\,dt$, $x\in\mathbb{R}$. Then, $\lim_{x\to 0}\dfrac{f(x)}{x^{3}}$ is equal to

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Limits

Let

f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t

x \in R

. Then,

lim_{x \to 0} \frac{f ( x )}{x ^{3}}

is equal to

Answer: (A)

f (x) = \int_{0}^{x} (t + sin (1 - e^{t})) d t

. Near

t = 0

1 - e^{t} = - t - \frac{t ^{2}}{2} - \frac{t ^{3}}{6} + O (t^{4})

, and

sin (1 - e^{t}) = (1 - e^{t}) - \frac{1}{6} (1 - e^{t})^{3} + O (t^{4}) = - t - \frac{t ^{2}}{2} + \frac{t ^{3}}{6} - \frac{t ^{3}}{6} \cdot (- 1) + O (t^{4})

.

More directly,

sin (1 - e^{t}) \approx - t - \frac{t ^{2}}{2} + O (t^{4})

(the cubic terms cancel), so the integrand

\approx t + (- t - \frac{t ^{2}}{2}) = - \frac{t ^{2}}{2}

f (x) \approx - \frac{x ^{3}}{6} \Rightarrow x \to 0 lim \frac{f ( x )}{x ^{3}} = - \frac{1}{6}

JEE Main · 2024 · 04 Apr 2024 · Shift 2 · Single correct option

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