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Question Number 168004 by mathlove last updated on 31/Mar/22

lim_(x→∞) (√(4x^2 −16x+1))−2x+3=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}−\mathrm{2}{x}+\mathrm{3}=? \\ $$

Answered by nurtani last updated on 31/Mar/22

lim_(x→∞)  (√(4x^2 −16x+1))−2x+3 =lim_(x→∞)  (√(4x^2 −16x+1))−(2x−3)  = lim_(x→∞)  (√(4x^2 −16x+1))−(√((2x−3)^2 ))  = lim_(x→∞)  (√(4x^2 −16x+1))−(√(4x^2 −12x+9))  = lim_(x→∞)  (√(4x^2 −16x+1))−(√(4x^2 −12x+9 ))× (((√(4x^2 −16x+1))+(√(4x^2 −12x+9)))/( (√(4x^2 −16x+1+))(√(4x^2 −12x+9))))  = lim_(x→∞)  ((4x^2 −16x+1−(4x^2 −12x+9))/( (√(4x^2 −16x+1))+(√(4x^2 −12x+9))))  = lim_(x→∞)  ((4x^2 −16x+1−4x^2 +12x−9)/( (√(4x^2 −16x+1))+(√(4x^2 −12x+9))))  = lim_(x→∞)  ((−4x−8)/( (√(4x^2 −16x+1))+(√(4x^2 −12x+9))))  = ((−4)/( (√4)+(√4)))=((−4)/(2+2))=((−4)/4)= −1

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}−\mathrm{2}{x}+\mathrm{3}\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}−\left(\mathrm{2}{x}−\mathrm{3}\right) \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}−\sqrt{\left(\mathrm{2}{x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}\:}×\:\frac{\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}+}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}−\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}\right)}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\cancel{\mathrm{4}{x}^{\mathrm{2}} }−\mathrm{16}{x}+\mathrm{1}−\cancel{\mathrm{4}{x}^{\mathrm{2}} }+\mathrm{12}{x}−\mathrm{9}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}}} \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{−\mathrm{4}{x}−\mathrm{8}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{16}{x}+\mathrm{1}}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{9}}} \\ $$$$=\:\frac{−\mathrm{4}}{\:\sqrt{\mathrm{4}}+\sqrt{\mathrm{4}}}=\frac{−\mathrm{4}}{\mathrm{2}+\mathrm{2}}=\frac{−\mathrm{4}}{\mathrm{4}}=\:−\mathrm{1} \\ $$$$ \\ $$

Commented by mathlove last updated on 31/Mar/22

thanks

$${thanks} \\ $$

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