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

lim_(x→0) ((∫_0 ^x^2   (√(4+t^3  )) dt)/x^2 ) ?

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\underset{\mathrm{0}} {\overset{{x}^{\mathrm{2}} } {\int}}\:\sqrt{\mathrm{4}+{t}^{\mathrm{3}} \:}\:{dt}}{{x}^{\mathrm{2}} }\:?\: \\ $$

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

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

let use hospital rule  we have lim_(x→0)  ((∫_0 ^x^2  (√(4+t^3 )))/x^2 )dt  =lim_(x→0)     ((2x(√(4+x^6 )))/(2x)) =lim_(x→0)  (√(4+x^6  ))=(√4)=2

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

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