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

Answered by aleks041103 last updated on 10/Jul/22

$$lny=\frac{ln\left(tgx\right)}{1+\sqrt[3]{1+{ln}^{2}x}}$$$$\Rightarrow L=\underset{x\to 0}{lim}ln\left(y\right)=\left[\frac{-\mathrm{\infty }}{\mathrm{\infty }}\right]$$$$\Rightarrow L^{\prime }Hopital$$$$L=\underset{x\to 0}{lim}\frac{\frac{1}{tg\left(x\right){cos}^2\left(x\right)}}{\frac{\frac{2ln\left(x\right)}{x}}{3{(1+{ln}^2\left(x\right))}^{2/3}}}=\frac{3}{2}\underset{x\to 0}{lim}\frac{x{(1+{ln}^{2}x)}^{2/3}}{ln\left(x\right)sin\left(x\right)cos\left(x\right)}=$$$$=\frac{3}{2}\left(\frac{1}{\underset{x\to 0}{lim}cos\left(x\right)}\right)\left(\frac{1}{\underset{x\to 0}{lim}\frac{sin\left(x\right)}{x}}\right)\left(\underset{x\to 0}{lim}\frac{{(1+{ln}^{2}x)}^{2/3}}{ln\left(x\right)}\right)=$$$$=\frac{3}{2}\underset{x\to 0}{lim}\frac{{(1+{ln}^{2}x)}^{2/3}}{ln\left(x\right)}=\frac{3}{2}\underset{x\to -\mathrm{\infty }}{lim}\frac{x^{4/3}{(1+\frac{1}{x^2})}^{2/3}}{x}=$$$$=\frac{3}{2}{\underset{x\to -\mathrm{\infty }}{limx}}^{1/3}{(1+0)}^{2/3}\to -\mathrm{\infty }$$$$\Rightarrow L\to -\mathrm{\infty }$$

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