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Question Number 105748 by ZiYangLee last updated on 31/Jul/20

lim_(x→∞) (((x−1)^(100) (6x+1)^(100) )/((3x+5)^(200) ))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{100}} \left(\mathrm{6x}+\mathrm{1}\right)^{\mathrm{100}} }{\left(\mathrm{3x}+\mathrm{5}\right)^{\mathrm{200}} }=? \\ $$

Answered by john santu last updated on 31/Jul/20

= (6^(100) /3^(200) ) = ((2^(100) .3^(100) )/3^(200) ) = ((2/3))^(100)

$$=\:\frac{\mathrm{6}^{\mathrm{100}} }{\mathrm{3}^{\mathrm{200}} }\:=\:\frac{\mathrm{2}^{\mathrm{100}} .\mathrm{3}^{\mathrm{100}} }{\mathrm{3}^{\mathrm{200}} }\:=\:\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{100}} \\ $$

Answered by mathmax by abdo last updated on 31/Jul/20

=lim_(x→∞)  ((x^(100) ×(6x)^(100) )/((3x)^(200) )) =lim_(x→∞)  (6^(100) /3^(200) ) (x^(200) /x^(200) )=(6^(100) /9^(100) ) =((2/3))^(100)

$$=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\frac{\mathrm{x}^{\mathrm{100}} ×\left(\mathrm{6x}\right)^{\mathrm{100}} }{\left(\mathrm{3x}\right)^{\mathrm{200}} }\:=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \:\frac{\mathrm{6}^{\mathrm{100}} }{\mathrm{3}^{\mathrm{200}} }\:\frac{\mathrm{x}^{\mathrm{200}} }{\mathrm{x}^{\mathrm{200}} }=\frac{\mathrm{6}^{\mathrm{100}} }{\mathrm{9}^{\mathrm{100}} }\:=\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{100}} \\ $$

Commented by ZiYangLee last updated on 31/Jul/20

thx

$$\mathrm{thx} \\ $$

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