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Question Number 170321 by amin96 last updated on 20/May/22

Answered by aleks041103 last updated on 21/May/22

$$a)$$$$\frac{2^x}{3^{x^2}}=2^x{3}^{-x^2}=e^{ln\left(2\right)x-ln\left(3\right)x^2}$$$$\underset{x\to \mathrm{\infty }}{lim}\frac{2^x}{3^{x^2}}={\underset{x\to \mathrm{\infty }}{lime}}^{ln\left(2\right)x-ln\left(3\right)x^2}=e^{-\mathrm{\infty }}=0$$$$b)$$$$\underset{x\to 0^{+}}{lim}\frac{e^{-3/x}}{x^2}=\left[\frac{0}{0}\right]=\underset{1/x\to \mathrm{\infty }}{lim}{\left(\frac{1}{x}\right)}^2{e}^{-3\left(1/x\right)}$$$$=\underset{y\to \mathrm{\infty }}{lim}\frac{y^2}{e^{3y}}=\underset{y\to \mathrm{\infty }}{lim}\frac{2y}{3e^{3y}}=\underset{y\to \mathrm{\infty }}{lim}\frac{2}{9e^{3y}}=0$$$$c)$$$$\underset{x\to \mathrm{\infty }}{lim}\frac{{ln}^{5}x}{x^2}={\left(\underset{x\to \mathrm{\infty }}{lim}\frac{lnx}{x^{2/5}}\right)}^5={\left(\underset{x\to \mathrm{\infty }}{lim}\frac{lnx}{x^{2/5}}\right)}^5=$$$$={\left(\frac{5}{2}\underset{x\to \mathrm{\infty }}{lim}\frac{ln\left(x^{2/5}\right)}{x^{2/5}}\right)}^5=\frac{5^5}{2^5}{\left(\underset{y\to \mathrm{\infty }}{lim}\frac{lny}{y}\right)}^5=$$$$=\frac{625\times 5}{32}{\left(\underset{y\to \mathrm{\infty }}{lim}\frac{1/y}{1}\right)}^5=0$$$$d)$$$$byanalogy\to 0$$

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