Question Number 173351 by cortano1 last updated on 10/Jul/22
Answered by aleks041103 last updated on 10/Jul/22
![ln y = ((ln(tgx))/(1+((1+ln^2 x))^(1/3) )) ⇒L=lim_(x→0) ln(y)=[((−∞)/∞)] ⇒L′Hopital L=lim_(x→0) ((1/(tg(x)cos^2 (x)))/(((2ln(x))/x)/(3(1+ln^2 (x))^(2/3) )))=(3/2)lim_(x→0) ((x(1+ln^2 x)^(2/3) )/(ln(x)sin(x)cos(x)))= =(3/2)((1/(lim_(x→0) cos(x))))((1/(lim_(x→0) ((sin(x))/x))))(lim_(x→0) (((1+ln^2 x)^(2/3) )/(ln(x))))= =(3/2)lim_(x→0) (((1+ln^2 x)^(2/3) )/(ln(x)))=(3/2)lim_(x→−∞) ((x^(4/3) (1+(1/x^2 ))^(2/3) )/x)= =(3/2)lim_(x→−∞) x^(1/3) (1+0)^(2/3) →−∞ ⇒L→−∞](https://www.tinkutara.com/question/Q173359.png)
$${ln}\:{y}\:=\:\frac{{ln}\left({tgx}\right)}{\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{1}+{ln}^{\mathrm{2}} {x}}} \\ $$$$\Rightarrow{L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{ln}\left({y}\right)=\left[\frac{−\infty}{\infty}\right] \\ $$$$\Rightarrow{L}'{Hopital} \\ $$$${L}=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\frac{\mathrm{1}}{{tg}\left({x}\right){cos}^{\mathrm{2}} \left({x}\right)}}{\frac{\frac{\mathrm{2}{ln}\left({x}\right)}{{x}}}{\mathrm{3}\left(\mathrm{1}+{ln}^{\mathrm{2}} \left({x}\right)\right)^{\mathrm{2}/\mathrm{3}} }}=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{x}\left(\mathrm{1}+{ln}^{\mathrm{2}} {x}\right)^{\mathrm{2}/\mathrm{3}} }{{ln}\left({x}\right){sin}\left({x}\right){cos}\left({x}\right)}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{cos}\left({x}\right)}\right)\left(\frac{\mathrm{1}}{\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{{sin}\left({x}\right)}{{x}}}\right)\left(\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{ln}^{\mathrm{2}} {x}\right)^{\mathrm{2}/\mathrm{3}} }{{ln}\left({x}\right)}\right)= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{ln}^{\mathrm{2}} {x}\right)^{\mathrm{2}/\mathrm{3}} }{{ln}\left({x}\right)}=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow−\infty} {{lim}}\frac{{x}^{\mathrm{4}/\mathrm{3}} \:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}/\mathrm{3}} }{{x}}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{x}\rightarrow−\infty} {{lim}x}^{\mathrm{1}/\mathrm{3}} \left(\mathrm{1}+\mathrm{0}\right)^{\mathrm{2}/\mathrm{3}} \rightarrow−\infty \\ $$$$\Rightarrow{L}\rightarrow−\infty \\ $$