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How to solve the limit $$\lim_{x\rightarrow \infty }\int_{0}^{x}\sin\frac{\pi }{t+x}\, \mathrm{d}t$$ need some help.

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    sub $(t+x)^{-1}=y$ and thinks should become clearer – 2017-01-14
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    Give some context and tell about you thoughts! – 2017-01-14

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Hint: Using the taylor series of $\sin x$, we have $$x-\frac{x^{3}}{6}<\sin x0$$ when $x\to\infty$, $\dfrac{\pi}{x+t}\to0$, so $$\int_{0}^{x}\frac{\pi }{t+x}\, \mathrm{d}t-\frac{\pi ^{3}}{6}\int_{0}^{x}\frac{1 }{\left (t+x \right )^{3}}\, \mathrm{d}t<\int_{0}^{x}\sin\frac{\pi }{t+x}\, \mathrm{d}t<\int_{0}^{x}\frac{\pi }{t+x}\, \mathrm{d}t$$ Now we only need to calculate $\displaystyle \int_{0}^{x}\frac{\pi }{t+x}\, \mathrm{d}t$ and $\displaystyle \int_{0}^{x}\frac{1 }{\left (t+x \right )^{3}}\, \mathrm{d}t$, then use the Squeeze theorem and you will get the answer.