Let g(x) = int_0^{|x|^{3/4}} t^{2/3} sin frac | vedclass

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(C) We are given $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$. Since the integrand $f(t) = t^{2/3} \sin \frac{1}{t}$ is bounded near $t=

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(C) We are given $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$. Since the integrand $f(t) = t^{2/3} \sin \frac{1}{t}$ is bounded near $t=0$ (because $|\sin(1/t)| \leq 1$,so $|f(t)| \leq t^{2/3}$),the integral exists. We want to evaluate $\lim_{x \rightarrow 0} \frac{g(x)}{x}$. Since $g(0) = \int_0^0 \dots = 0$,this is a $0/0$ form. Using the squeeze theorem or direct estimation: $|g(x)| = \left| \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt \right| \leq \int_0^{|x|^{3/4}} |t^{2/3} \sin \frac{1}{t}| \, dt \leq \int_0^{|x|^{3/4}} t^{2/3} \, dt$. Evaluating the integral: $\int_0^{|x|^{3/4}} t^{2/3} \, dt = \left[ \frac{t^{5/3}}{5/3} \right]_0^{|x|^{3/4}} = \frac{3}{5} (|x|^{3/4})^{5/3} = \frac{3}{5} |x|^{5/4}$. Thus,$\left| \frac{g(x)}{x} \right| \leq \frac{\frac{3}{5} |x|^{5/4}}{|x|} = \frac{3}{5} |x|^{1/4}$. As $x \rightarrow 0$,$\frac{3}{5} |x|^{1/4} \rightarrow 0$. Therefore,by the squeeze theorem,$\lim_{x \rightarrow 0} \frac{g(x)}{x} = 0$.

Let $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$,for all real $x$. Then,$\lim_{x \rightarrow 0} \frac{g(x)}{x}$ is equal to

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