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　次の問いに答えよ．
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\item[(1)]$t>0$のとき
\[ -\frac{1}{t}<\int_t^{2t}\frac{\sin x}{x^2}dx<\frac{1}{t} \]
が成り立つことを示せ．
\item[(2)]$\displaystyle\lim_{t\to\infty}\int_t^{2t}\frac{\cos x}{x}dx=0$を示せ．
\item[(3)]$\displaystyle f(x)=\sin\left(\frac{3x}{2}\right)\sin\left(\frac{x}{2}\right)$とおく．
\[ \lim_{t\to\infty}\int_1^t\frac{f(x)}{x}dx=\frac{1}{2}\int_1^2\frac{\cos x}{x}dx \]
を示せ．
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