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Question Number 94723 by john santu last updated on 20/May/20

$$\underset{x\to 0}{\lim }\frac{\underset{0}{\stackrel{x^2}{\int }}\sqrt{4+t^3}dt}{x^2}?$$

Answered by i jagooll last updated on 20/May/20

Answered by mathmax by abdo last updated on 20/May/20

$$\mathrm{let}\mathrm{use}\mathrm{hospital}\mathrm{rule}\mathrm{we}\mathrm{have}{\lim }_{\mathrm{x}\to 0}\frac{{\int }_0^{{\mathrm{x}}^2}\sqrt{4+{\mathrm{t}}^3}}{{\mathrm{x}}^2}\mathrm{dt}$$$$={\lim }_{\mathrm{x}\to 0}\frac{2\mathrm{x}\sqrt{4+{\mathrm{x}}^6}}{2\mathrm{x}}={\lim }_{\mathrm{x}\to 0}\sqrt{4+{\mathrm{x}}^6}=\sqrt{4}=2$$

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