# Question and Answers Forum

Limits Questions

Question Number 101610 by bramlex last updated on 03/Jul/20

Answered by john santu last updated on 04/Jul/20

$$\mathrm{L}'\mathrm{Hopital} \\$$$$\Leftrightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} \sqrt{{x}^{\mathrm{3}} +\mathrm{5}}}{\mathrm{sin}\:\mathrm{2}{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{\mathrm{sin}\:\mathrm{2}{x}}\:×\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{x}\sqrt{{x}^{\mathrm{3}} +\mathrm{5}} \\$$$$=\:\mathrm{0}\:\left(\mathrm{JS}\circledast\right) \\$$

Answered by Dwaipayan Shikari last updated on 04/Jul/20

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{{x}} \mathrm{2}{u}^{\mathrm{2}} {du}}{{sin}^{\mathrm{2}} {x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}}{\mathrm{9}}\left[{u}^{\mathrm{3}} \right]_{\sqrt{\mathrm{5}}} ^{\sqrt{{x}^{\mathrm{3}} +\mathrm{5}}} }{{sin}^{\mathrm{2}} {x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}}{\mathrm{9}}\left[\left({x}^{\mathrm{3}} +\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{5}^{\frac{\mathrm{2}}{\mathrm{3}}} \right]}{{x}^{\mathrm{2}} {sin}^{\mathrm{2}} {x}}.{x}^{\mathrm{2}} \\$$$$\\$$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\mathrm{2}}{\mathrm{9}}\left[\left({x}^{\mathrm{3}} +\mathrm{5}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} −\mathrm{5}^{\frac{\mathrm{2}}{\mathrm{3}}} \right]}{\left({x}^{\mathrm{3}} +\mathrm{5}\right)−\mathrm{5}}.\frac{{x}^{\mathrm{2}} {x}}{{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{9}}.\frac{\mathrm{2}}{\mathrm{3}}\:\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{3}}} .{x}=\:\:\:\mathrm{0}\:\:\left({DS}\right)\:\:\blacksquare\bigstar\left\{\mathrm{assuming}\:\mathrm{sinx}=\mathrm{x}\right\} \\$$$$\\$$