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Question Number 146263 by bemath last updated on 12/Jul/21

  lim_(x→−∞) (−2x+1−(√(4x^2 −5x+8)))=?

$$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(−\mathrm{2}{x}+\mathrm{1}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}}\right)=? \\ $$

Answered by gsk2684 last updated on 12/Jul/21

lim_(x→−∞) (((−2x+1)^2 −(4x^2 −5x+8))/(−2x+1+(√(4x^2 −5x+8))))  lim_(x→−∞) ((x−7)/(−2x+1−x(√(4−(5/x)+(8/x^2 )))))  (1/(−2−(√4)))=−(1/4)

$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\left(−\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} −\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}\right)}{−\mathrm{2}{x}+\mathrm{1}+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{8}}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}−\mathrm{7}}{−\mathrm{2}{x}+\mathrm{1}−{x}\sqrt{\mathrm{4}−\frac{\mathrm{5}}{{x}}+\frac{\mathrm{8}}{{x}^{\mathrm{2}} }}} \\ $$$$\frac{\mathrm{1}}{−\mathrm{2}−\sqrt{\mathrm{4}}}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Commented by mr W last updated on 12/Jul/21

correct

$${correct} \\ $$

Answered by iloveisrael last updated on 13/Jul/21

 lim_(x→−∞) (−2x+1−(√(4x^2 −5x+8)))=?  solution :    lim_(x→−∞) (−2x+1−(√(4x^2 −5x+8)) )  = lim_(x→−∞) (−2x+1−(√4)∣x−(5/(2.4))∣)  =lim_(x→−∞) (−2x+1−2(−x+(5/8)))  =lim_(x→−∞) (−2x+1+2x−(5/4))  =lim_(x→−∞) (1−(5/4))=−(1/4).✓

$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(−\mathrm{2x}+\mathrm{1}−\sqrt{\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{8}}\right)=? \\ $$$$\mathrm{solution}\::\: \\ $$$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(−\mathrm{2x}+\mathrm{1}−\sqrt{\mathrm{4x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{8}}\:\right) \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(−\mathrm{2x}+\mathrm{1}−\sqrt{\mathrm{4}}\mid\mathrm{x}−\frac{\mathrm{5}}{\mathrm{2}.\mathrm{4}}\mid\right) \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(−\mathrm{2x}+\mathrm{1}−\mathrm{2}\left(−\mathrm{x}+\frac{\mathrm{5}}{\mathrm{8}}\right)\right) \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(−\mathrm{2x}+\mathrm{1}+\mathrm{2x}−\frac{\mathrm{5}}{\mathrm{4}}\right) \\ $$$$=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{5}}{\mathrm{4}}\right)=−\frac{\mathrm{1}}{\mathrm{4}}.\checkmark \\ $$

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