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Question Number 32044 by abdo imad last updated on 18/Mar/18

$$fimd{lim}_{x\to 0}\frac{1}{x^3}{\int }_0^x{t}^{2}ln(1+sint)dt.$$

Commented by abdo imad last updated on 20/Mar/18

$$ln(1+sint)\sim ln(1+t)\sim t\Rightarrow t^{2}ln(1+sint)\sim t^3\Rightarrow$$$${\int }_0^x{t}^{2}ln(1+sint)dt\sim {\int }_0^x{t}^{3}dt=\frac{x^4}{4}\Rightarrow$$$${lim}_{x\to 0}{\int }_0^x{t}^{2}ln(1+sint)dt={lim}_{x\to 0}\frac{x}{4}=0$$

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