Q40 / 90 · JEE Main 2024 (04 Apr Shift 2) · Mathematics
Let $f(x)=\int_0^x\left(t+\sin \left(1-e^{\prime}\right)\right) d t, x \in \mathbb{R}$. Then, $\lim _{x \rightarrow 0} \frac{f(x)}{x^3}$ is equal to
- Correct answerA$-\frac{1}{6}$
- B$\frac{2}{3}$
- C$-\frac{2}{3}$
- D$\frac{1}{6}$
Solution
$\lim _{x \rightarrow 0} \frac{f(x)}{x^3}$
Using L Hopital Rule. $\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}$ (Again L Hopital) Using L.H. Rule $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} \\ & =-\frac{1}{6} \end{aligned}$
Using L Hopital Rule. $\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}$ (Again L Hopital) Using L.H. Rule $\begin{aligned} & =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} \\ & =-\frac{1}{6} \end{aligned}$
Hence, the correct answer is option: 'A'.