Question

$\lim_{x\to\infty} \frac{\int^{2x}_{0 } xe^{x^2}dx}{e^{4x^2}} $ equals

• A. 0
• B. $\infty$
• C. 2
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

Consider $\displaystyle\lim _{x \rightarrow \infty} \frac{\int\limits_{0}^{2 x} x e^{x^{2}} d x}{e^{4 x^{2}}}$
$= \displaystyle\lim _{x \rightarrow \infty} \frac{2 \int\limits_{0}^{2 x} x e^{x^{2}} d x}{2 e^{4 x^{2}}}$
$= \displaystyle\lim _{x \rightarrow \infty} \frac{2 \int\limits_{0}^{2 x} e^{x^{2}} d\left(x^{2}\right)}{2 e^{4 x^{2}}}$
$= \displaystyle\lim _{x \rightarrow \infty} \frac{\left[e^{x^{2}}\right]_{0}^{2 x}}{e^{4 x^{2}}}$
$= \displaystyle\lim _{x \rightarrow \infty} \frac{e^{x^{2}}-1}{2 e^{4 x^{2}}}$
$= \displaystyle\lim _{x \rightarrow \infty}\left(\frac{1}{2}-\frac{1}{2 e^{4 x^{2}}}\right)$
$=\frac{1}{2}$