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Author: Tinku Tara

Question-69710

Question Number 69710 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Commented by mathmax by abdo last updated on 27/Sep/19 $${we}\:{have}\:{x}^{\mathrm{2}} −\mathrm{6}{x}\:+\mathrm{5}\:={x}^{\mathrm{2}} −{x}\:−\mathrm{5}\left({x}−\mathrm{1}\right)={x}\left({x}−\mathrm{1}\right)−\mathrm{5}\left({x}−\mathrm{1}\right) \\ $$$$=\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}}…

Question-69709

Question Number 69709 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Commented by kaivan.ahmadi last updated on 26/Sep/19 $${lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}{x}}{\mathrm{3}^{{x}} {ln}\mathrm{3}}={lim}_{{x}\rightarrow\infty} \frac{\mathrm{2}}{\mathrm{3}^{{x}} \left({ln}\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ Commented…

Question-4166

Question Number 4166 by Filup last updated on 30/Dec/15 Commented by prakash jain last updated on 31/Dec/15 $$\mathrm{To}\:\mathrm{go}\:\mathrm{broke}:\:{k}\:\mathrm{wins},\:\mathrm{2}+{k}\:\mathrm{losses} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{k}} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}+{k}} =\:\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{16}}}\:=\:\frac{\mathrm{1}}{\mathrm{13}}\:?…