Question

The limit $\displaystyle\lim _{x \rightarrow \infty} x^{2} \int\limits_{0}^{x} e^{t^{3}-x^{3}} d t$ equals

• A. $\frac{1}{3}$
• B. $2$
• C. $\infty$
• D. $\frac{2}{3}$

Answer 1

Answer: (A)

We have,
$\displaystyle\lim_{x\to\infty} x^{2}\int\limits_{0}^{x} e^{t^3 -x^3} dt $
$=\displaystyle\lim_{x\to\infty} x^{2}e^{-x^3} \int\limits_{0}^{x} e^{t^3}dt $
$=\displaystyle\lim_{x\to\infty} \frac{x^{2}\int\limits_{0}^{x}e^{t^3}dt}{e^{x^3}}$
Apply L Hospital’s rule
$=\displaystyle\lim_{x\to\infty} \frac{2x\int\limits_{0}^{x}e^{t^3}dt+x^{2}e^{x^3}}{3x^{2}e^{x^3}}$
$=\displaystyle\lim_{x\to\infty} \frac{2\int\limits_{0}^{x}e^{t^3}dt+x^{2}e^{x^3}}{3x^{2} e^{x^3}}$
Again Apply L Hospital’s rule, we get
$=\displaystyle\lim_{x\to\infty} \frac{2e^{x^3}+e^{x^3}+3x^{3}e^{x^3}}{3e^{x^3}+9x^{3}ex^{3}}$
$=\displaystyle\lim_{x\to\infty} \frac{e^{x^3}\left(3+3x^{3}\right)}{ex^{3}\left(3+9x^{3}\right)}=\frac{1}{3}$