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JEE Main 2024
04-04-2024 S2
Question
Let $f(x)=\int_0^x\left(t+\sin \left(1-e^t\right)\right) d t, x \in \mathbb{R}$.
Then $\lim _{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to
Select the correct option:
A
$\frac{1}{6}$
B
$-\frac{1}{6}$
C
$-\frac{2}{3}$
D
$\frac{2}{3}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
View Solution
Solution
$ \lim _{x \rightarrow 0} \frac{f(x)}{x^3} $
Using L Hopital Rule.
$ \lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2} \text { (Again L Hopital) } $
Using L.H. Rule
$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} \\ & =-\frac{1}{6} \end{aligned} $
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