Let f ( x ) = 0 x ( t + sin ( 1 - e t ) ) d t , ? x R . Then, lim x 0 f ( x ) x 3 is equal to [2024]

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Solution

Q.
$Let f (x) = \int_{0}^{x} (t + \sin (1 - e^{t})) d t, x \in R . Then, \lim_{x \to 0} \frac{f (x)}{x^{3}} is equal to$ [2024]

1 $\frac{1}{6}$

2 $- \frac{1}{6}$

3 $\frac{2}{3}$

4 $- \frac{2}{3}$

Ans.
(2)

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

$Now, \lim_{x \to 0} \frac{f (x)}{x^{3}} (\frac{0}{0} form) = \lim_{x \to 0} \frac{f' (x)}{3 x^{2}}$

$= \lim_{x \to 0} \frac{x + \sin (1 - e^{x})}{3 x^{2}} (\frac{0}{0} form)$

$= \lim_{x \to 0} \frac{1 + \cos (1 - e^{x}) (- e^{x})}{6 x} (\frac{0}{0} form)$

$= \lim_{x \to 0} \frac{- \sin (1 - e^{x}) {(e^{x})}^{2} + \cos (1 - e^{x}) (- e^{x})}{6} = - \frac{1}{6}$

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