Question Number 170321 by amin96 last updated on 20/May/22
Answered by aleks041103 last updated on 21/May/22
![a) (2^x /3^x^2 )=2^x 3^(−x^2 ) =e^(ln(2)x−ln(3)x^2 ) lim_(x→∞) (2^x /3^x^2 )=lim_(x→∞) e^(ln(2)x−ln(3)x^2 ) =e^(−∞) =0 b) lim_(x→0^+ ) (e^(−3/x) /x^2 )=[(0/0)]=lim_(1/x→∞) ((1/x))^2 e^(−3(1/x)) =lim_(y→∞) (y^2 /e^(3y) )=lim_(y→∞) ((2y)/(3e^(3y) ))=lim_(y→∞) (2/(9e^(3y) ))=0 c) lim_(x→∞) ((ln^5 x)/x^2 )=(lim_(x→∞) ((lnx)/x^(2/5) ))^5 =(lim_(x→∞) ((lnx)/x^(2/5) ))^5 = =((5/2)lim_(x→∞) ((ln(x^(2/5) ))/x^(2/5) ))^5 =(5^5 /2^5 )(lim_(y→∞) ((lny)/y))^5 = =((625×5)/(32))(lim_(y→∞) ((1/y)/1))^5 =0 d) by analogy→0](https://www.tinkutara.com/question/Q170347.png)
$$\left.{a}\right) \\ $$$$\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}^{\mathrm{2}} } }=\mathrm{2}^{{x}} \mathrm{3}^{−{x}^{\mathrm{2}} } ={e}^{{ln}\left(\mathrm{2}\right){x}−{ln}\left(\mathrm{3}\right){x}^{\mathrm{2}} } \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{2}^{{x}} }{\mathrm{3}^{{x}^{\mathrm{2}} } }=\underset{{x}\rightarrow\infty} {{lim}e}^{{ln}\left(\mathrm{2}\right){x}−{ln}\left(\mathrm{3}\right){x}^{\mathrm{2}} } ={e}^{−\infty} =\mathrm{0} \\ $$$$\left.{b}\right) \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{{e}^{−\mathrm{3}/{x}} }{{x}^{\mathrm{2}} }=\left[\frac{\mathrm{0}}{\mathrm{0}}\right]=\underset{\mathrm{1}/{x}\rightarrow\infty} {{lim}}\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} {e}^{−\mathrm{3}\left(\mathrm{1}/{x}\right)} \\ $$$$=\underset{{y}\rightarrow\infty} {{lim}}\frac{{y}^{\mathrm{2}} }{{e}^{\mathrm{3}{y}} }=\underset{{y}\rightarrow\infty} {{lim}}\frac{\mathrm{2}{y}}{\mathrm{3}{e}^{\mathrm{3}{y}} }=\underset{{y}\rightarrow\infty} {{lim}}\frac{\mathrm{2}}{\mathrm{9}{e}^{\mathrm{3}{y}} }=\mathrm{0} \\ $$$$\left.{c}\right) \\ $$$$\underset{{x}\rightarrow\infty} {{lim}}\frac{{ln}^{\mathrm{5}} {x}}{{x}^{\mathrm{2}} }=\left(\underset{{x}\rightarrow\infty} {{lim}}\frac{{lnx}}{{x}^{\mathrm{2}/\mathrm{5}} }\right)^{\mathrm{5}} =\left(\underset{{x}\rightarrow\infty} {{lim}}\frac{{lnx}}{{x}^{\mathrm{2}/\mathrm{5}} }\right)^{\mathrm{5}} = \\ $$$$=\left(\frac{\mathrm{5}}{\mathrm{2}}\underset{{x}\rightarrow\infty} {{lim}}\frac{{ln}\left({x}^{\mathrm{2}/\mathrm{5}} \right)}{{x}^{\mathrm{2}/\mathrm{5}} }\right)^{\mathrm{5}} =\frac{\mathrm{5}^{\mathrm{5}} }{\mathrm{2}^{\mathrm{5}} }\left(\underset{{y}\rightarrow\infty} {{lim}}\frac{{lny}}{{y}}\right)^{\mathrm{5}} = \\ $$$$=\frac{\mathrm{625}×\mathrm{5}}{\mathrm{32}}\left(\underset{{y}\rightarrow\infty} {{lim}}\frac{\mathrm{1}/{y}}{\mathrm{1}}\right)^{\mathrm{5}} =\mathrm{0} \\ $$$$\left.{d}\right) \\ $$$${by}\:{analogy}\rightarrow\mathrm{0} \\ $$