limy→∞log1+ye2ylog1+ye3y-log y=

limy→∞log1+ye2y-log ylog1+ye3y-log y =limy→∞log1y+e2ylog1y+e3y 00 form =limy→∞11y+e2y-1y2+e2y-2y211y+e3y-1y2+e3y-3y2 =limy→∞11y+e2y1+e2y.211y+e3y1+e3y.3 =1+21+3=34

limy→∞log1+ye2ylog1+ye3y-log y=

limy→∞log1+ye2y-log ylog1+ye3y-log y =limy→∞log1y+e2ylog1y+e3y 00 form =limy→∞11y+e2y-1y2+e2y-2y211y+e3y-1y2+e3y-3y2 =limy→∞11y+e2y1+e2y.211y+e3y1+e3y.3 =1+21+3=34

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$\lim_{y \to \infty} \frac{\log (1 + {ye}^{\frac{2}{y}})}{\log (1 + {ye}^{\frac{3}{y}}) - \log y} =$

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$\frac{1}{2}$

$\frac{2}{3}$

$\frac{3}{4}$

$\frac{4}{5}$

answer is C.

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Detailed Solution

$\lim_{y \to \infty} \frac{\log (1 + {ye}^{\frac{2}{y}}) - \log y}{\log (1 + {ye}^{\frac{3}{y}}) - \log y} = \lim_{y \to \infty} \frac{\log (\frac{1}{y} + e^{\frac{2}{y}})}{\log (\frac{1}{y} + e^{\frac{3}{y}})} (\frac{0}{0} form) = \lim_{y \to \infty} \frac{\frac{1}{\frac{1}{y} + e^{\frac{2}{y}}} (- \frac{1}{y^{2}} + e^{\frac{2}{y}} (\frac{- 2}{y^{2}}))}{\frac{1}{\frac{1}{y} + e^{\frac{3}{y}}} (- \frac{1}{y^{2}} + e^{\frac{3}{y}} (\frac{- 3}{y^{2}}))} = \lim_{y \to \infty} \frac{\frac{1}{\frac{1}{y} + e^{\frac{2}{y}}} (1 + e^{\frac{2}{y}} . 2)}{\frac{1}{\frac{1}{y} + e^{\frac{3}{y}}} (1 + e^{\frac{3}{y}} . 3)} = \frac{1 + 2}{1 + 3} = \frac{3}{4}$