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Question Number 99348 by bobhans last updated on 20/Jun/20

Commented by bemath last updated on 20/Jun/20

lim_(x→−∞) ((3x+(√x^2 ))/(7x−5(√x^2 ))) = lim_(x→−∞) ((3x−x)/(7x+5x))  = lim_(x→−∞) ((2x)/(12x)) = (1/6)

$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3x}+\sqrt{\mathrm{x}^{\mathrm{2}} }}{\mathrm{7x}−\mathrm{5}\sqrt{\mathrm{x}^{\mathrm{2}} }}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{3x}−\mathrm{x}}{\mathrm{7x}+\mathrm{5x}} \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{2x}}{\mathrm{12x}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Answered by abdomsup last updated on 20/Jun/20

=lim_(x→−∞)  ((3x−x)/(7x+5x))  =lim_(x→−∞)  ((2x)/(12x)) =(1/6)

$$={lim}_{{x}\rightarrow−\infty} \:\frac{\mathrm{3}{x}−{x}}{\mathrm{7}{x}+\mathrm{5}{x}} \\ $$$$={lim}_{{x}\rightarrow−\infty} \:\frac{\mathrm{2}{x}}{\mathrm{12}{x}}\:=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

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