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$$\lim _{x \rightarrow \infty} \frac{\int_{0}^{2 x} x e^{x^{2}} d x}{e^{4 x^{2}}} \text { equals }$$
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Verified Answer
The correct answer is:
$\frac{1}{2}$
Consider $\lim _{x \rightarrow \infty} \frac{\int_{0}^{2 x} x e^{x^{2}} d x}{e^{4 x^{2}}}$
$$
\begin{array}{l}
=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{2 x} x e^{x^{2}} d x}{2 e^{4 x^{2}}} \\
=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{2 x} e^{x^{2}} d\left(x^{2}\right)}{2 e^{4 x^{2}}} \\
=\lim _{x \rightarrow \infty} \frac{\left[e^{x^{2}}\right]_{0}^{2 x}}{e^{4 x^{2}}}=\lim _{x \rightarrow \infty} \frac{e^{4 x^{2}}-1}{2 e^{4 x^{2}}} \\
=\lim _{x \rightarrow \infty}\left(\frac{1}{2}-\frac{1}{2 e^{4 x^{2}}}\right)=\frac{1}{2}
\end{array}
$$
$$
\begin{array}{l}
=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{2 x} x e^{x^{2}} d x}{2 e^{4 x^{2}}} \\
=\lim _{x \rightarrow \infty} \frac{2 \int_{0}^{2 x} e^{x^{2}} d\left(x^{2}\right)}{2 e^{4 x^{2}}} \\
=\lim _{x \rightarrow \infty} \frac{\left[e^{x^{2}}\right]_{0}^{2 x}}{e^{4 x^{2}}}=\lim _{x \rightarrow \infty} \frac{e^{4 x^{2}}-1}{2 e^{4 x^{2}}} \\
=\lim _{x \rightarrow \infty}\left(\frac{1}{2}-\frac{1}{2 e^{4 x^{2}}}\right)=\frac{1}{2}
\end{array}
$$
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