Limits Questions

Question Number 173351 by cortano1 last updated on 10/Jul/22

Answered by aleks041103 last updated on 10/Jul/22

$${ln}\:{y}\:=\:\frac{{ln}\left({tgx}\right)}{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{1}+{ln}^{\mathrm{2}} {x}}} \\$$$$\Rightarrow{L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{ln}\left({y}\right)=\left[\frac{−\infty}{\infty}\right] \\$$$$\Rightarrow{L}'{Hopital} \\$$$${L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{\mathrm{1}}{{tg}\left({x}\right){cos}^{\mathrm{2}} \left({x}\right)}}{\frac{\frac{\mathrm{2}{ln}\left({x}\right)}{{x}}}{\mathrm{3}\left(\mathrm{1}+{ln}^{\mathrm{2}} \left({x}\right)\right)^{\mathrm{2}/\mathrm{3}} }}=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\left(\mathrm{1}+{ln}^{\mathrm{2}} {x}\right)^{\mathrm{2}/\mathrm{3}} }{{ln}\left({x}\right){sin}\left({x}\right){cos}\left({x}\right)}= \\$$$$=\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{cos}\left({x}\right)}\right)\left(\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({x}\right)}{{x}}}\right)\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{ln}^{\mathrm{2}} {x}\right)^{\mathrm{2}/\mathrm{3}} }{{ln}\left({x}\right)}\right)= \\$$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{ln}^{\mathrm{2}} {x}\right)^{\mathrm{2}/\mathrm{3}} }{{ln}\left({x}\right)}=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow−\infty} {{lim}}\frac{{x}^{\mathrm{4}/\mathrm{3}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}/\mathrm{3}} }{{x}}= \\$$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow−\infty} {{lim}x}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{2}/\mathrm{3}} \rightarrow−\infty \\$$$$\Rightarrow{L}\rightarrow−\infty \\$$