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Let $g(x)=\int_{0}^{|x|^{2 / 4}} t^{2 / 3} \sin \frac{1}{t} d t$, for all real $x$. Then $\lim _{x \rightarrow 0} \frac{\mathrm{g}(\mathrm{x})}{\mathrm{x}}$ is equal to
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The correct answer is:
$\infty$
Apply L hospital rule
$\begin{array}{l}
g^{\prime}(x)=|x|^{1 / 2} \sin \left(\frac{1}{|x|^{3 / 4}}\right) \\
\lim _{x \rightarrow 0} g^{\prime}(X) \\
\lim _{x \rightarrow 0}|x|^{1 / 2} \sin \left(\frac{1}{|x|^{3 / 4}}\right)=0
\end{array}$
$\begin{array}{l}
g^{\prime}(x)=|x|^{1 / 2} \sin \left(\frac{1}{|x|^{3 / 4}}\right) \\
\lim _{x \rightarrow 0} g^{\prime}(X) \\
\lim _{x \rightarrow 0}|x|^{1 / 2} \sin \left(\frac{1}{|x|^{3 / 4}}\right)=0
\end{array}$
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