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Question Number 168009 by mathlove last updated on 31/Mar/22

lim_(x→∞) (((√(4x^2 −4x+1))+3x)/( (√(x^2 +x−5))+x))=?

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

Answered by nurtani last updated on 31/Mar/22

lim_(x→∞) (((√(4x^2 −4x+1))+3x)/( (√(x^2 +x−5))+x))=lim_(x→∞) ((((√(4x^2 −4x+1))/x)+((3x)/x))/( ((√(x^2 +x−5))/x)+(x/x))) =lim_(x→∞) (((√(((4x^2 )/x^2 )−((4x)/x^2 )+(1/x^2 )))+((3x)/x))/( (√((x^2 /x^2 )+(x/x^2 )−(5/x^2 )))+(x/x))) = lim_(x→∞) (((√(4−0+0))+3)/( (√(1+0−0))+1)) = (((√4)+3)/( (√1)+1)) = ((2+3)/(1+1)) = (5/2)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}+\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{5}}+{x}}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}}{{x}}+\frac{\mathrm{3}{x}}{{x}}}{\:\frac{\sqrt{{x}^{\mathrm{2}} +{x}−\mathrm{5}}}{{x}}+\frac{{x}}{{x}}} \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\frac{\mathrm{4}{x}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}+\frac{\mathrm{3}{x}}{{x}}}{\:\sqrt{\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{{x}}{{x}^{\mathrm{2}} }−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}+\frac{{x}}{{x}}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}−\mathrm{0}+\mathrm{0}}+\mathrm{3}}{\:\sqrt{\mathrm{1}+\mathrm{0}−\mathrm{0}}+\mathrm{1}}\:=\:\frac{\sqrt{\mathrm{4}}+\mathrm{3}}{\:\sqrt{\mathrm{1}}+\mathrm{1}}\:=\:\frac{\mathrm{2}+\mathrm{3}}{\mathrm{1}+\mathrm{1}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by peter frank last updated on 02/Apr/22

thanks

$$\mathrm{thanks} \\ $$

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