Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 171385 by cortano1 last updated on 14/Jun/22

lim_(x→∞) ((rx^5 −4x^3 +3)/(sx^5 −2x+1)) = (1/4) 2r+3s=?

$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{rx}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}}{{sx}^{\mathrm{5}} −\mathrm{2}{x}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\:\:\mathrm{2}{r}+\mathrm{3}{s}=? \\ $$

Answered by qaz last updated on 14/Jun/22

lim_(x→∞) ((rx^5 −4x^3 +3)/(sx^5 −2x+1))=(r/s)=(1/4) ⇒ 2r+3s= ???

$$\underset{{x}\rightarrow\infty} {{lim}}\frac{{rx}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}}{{sx}^{\mathrm{5}} −\mathrm{2}{x}+\mathrm{1}}=\frac{{r}}{{s}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{2}{r}+\mathrm{3}{s}=\:??? \\ $$

Answered by nurtani last updated on 14/Jun/22

14 ???

$$\mathrm{14}\:??? \\ $$

Answered by alephzero last updated on 14/Jun/22

lim_(x→∞) ((rx^5 −4x^3 +3)/(sx^5 −2x+1)) = lim_(x→∞) ((r5x^4 −12x^2 )/(s5x^4 −2x)) = lim_(x→∞) ((x^2 (r5x^2 −12))/(x(s5x^3 −2))) = lim_(x→∞) ((r5x^3 −12x)/(s5x^3 −2x)) = lim_(x→∞) ((r15x^2 )/(s15x^2 )) = lim_(x→∞) (r/s) = (r/s) = (1/4) ⇒ r = k ∧ s = 4r = 4k, k ∈ Z ⇒ 2r+3s = 2k+3(4k) = 2k+12k = = 14k

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{rx}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{3}}{{sx}^{\mathrm{5}} −\mathrm{2}{x}+\mathrm{1}}\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{r}\mathrm{5}{x}^{\mathrm{4}} −\mathrm{12}{x}^{\mathrm{2}} }{{s}\mathrm{5}{x}^{\mathrm{4}} −\mathrm{2}{x}}\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} \left({r}\mathrm{5}{x}^{\mathrm{2}} −\mathrm{12}\right)}{{x}\left({s}\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}\right)}\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{r}\mathrm{5}{x}^{\mathrm{3}} −\mathrm{12}{x}}{{s}\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}{x}}\:= \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{r}\mathrm{15}{x}^{\mathrm{2}} }{{s}\mathrm{15}{x}^{\mathrm{2}} }\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{r}}{{s}}\:=\:\frac{{r}}{{s}}\:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:{r}\:=\:{k}\:\wedge\:{s}\:=\:\mathrm{4}{r}\:=\:\mathrm{4}{k},\:\:{k}\:\in\:\mathbb{Z} \\ $$$$\Rightarrow\:\mathrm{2}{r}+\mathrm{3}{s}\:=\:\mathrm{2}{k}+\mathrm{3}\left(\mathrm{4}{k}\right)\:=\:\mathrm{2}{k}+\mathrm{12}{k}\:= \\ $$$$=\:\mathrm{14}{k} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com