Explanation

(d)
$\lim_{x \rightarrow \infty}\frac{\int_{0}^{2x}{xe^{x^{2}}dx}}{e^{4x^{2}}} = \lim_{x \rightarrow \infty}\frac{\int_{0}^{2x}{e^{x^{2}}d(x)^{2}}}{2e^{4x^{2}}}$
$= \lim_{x \rightarrow \infty}\frac{\left\lbrack e^{x^{2}} \right\rbrack_{0}^{2x}}{2e^{4x^{2}}}$
$= \lim_{x \rightarrow \infty}\frac{e^{4x^{2}} - 1}{2e^{4x^{2}}}$
$= \lim_{x \rightarrow \infty}\left( \frac{1}{2} - \frac{1}{e^{4x^{2}}} \right) = \frac{1}{2}$