MCQ
The limit $\lim _{x \rightarrow \infty} x^2 \int \limits_0^x e^{t^3-x^3} d t$ equals
- ✓$\frac{1}{3}$
- B$2$
- C$\infty$
- D$\frac{2}{3}$
We have,
$\lim _{x \rightarrow \infty} x^2 \int \limits_0^x e^{t^3-x^3} d t =\lim _{x \rightarrow \infty} x^2 e^{-x^3} \int \limits_0^x e^{e^3} d t$
$=\lim _{x \rightarrow \infty} \frac{x^2 \int \limits_0^x e^{t^3} d t}{e^{x^3}}$
Apply $L$ Hospital's rule
$=\lim _{x \rightarrow \infty} \frac{2 x \int \limits_0^x e^{t^3} d t+x^2 e^{x^3}}{3 x^2 e^{x^3}}$ $=\lim _{x \rightarrow \infty} \frac{2 \int \limits_0^x e^{t^3} d t+x^2 e^{x^3}}{3 x^2 e^{x^3}}$
Again Apply $L$ Hospital's rule, we get
$=\lim _{x \rightarrow \infty} \frac{2 e^{x^3}+e^{x^3}+3 x^3 e^{x^3}}{3 e^{x^3}+9 x^3 e^{x^3}}$
$=\lim _{x \rightarrow \infty} \frac{e^{x^3}\left(3+3 x^3\right)}{e^{x^3}\left(3+9 x^3\right)}=\frac{1}{3}$
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If
$\frac{\text{d}}{\text{dx}}\text{f}\text{(x)}=4\text{x}^3-\frac{3}{\text{x}^4}$such that $\text{f}(2)=0.$Then $\text{f}\text{(x)}$ is