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lim-x-2x-3x-2-3x-x-1-




Question Number 144558 by imjagoll last updated on 26/Jun/21
  lim_(x→−∞) (((2x−(√(3x^2 +3x)))/(x−1)))=?
$$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}}}{\mathrm{x}−\mathrm{1}}\right)=? \\ $$
Answered by liberty last updated on 26/Jun/21
  lim_(x→−∞) (((2x−(√(3x^2 +3x)))/(x−1)))=?  Sol :    lim_(x→−∞) (((2x−(√(3x^2 +3x)))/(x−1)))  = lim_(x→−∞) (((2x−∣x∣(√(3+(3/x))))/(x−1)))  = lim_(x→−∞) ((x(2+(√(3+3x^(−1) )))/(x(1−x^(−1) )))  = lim_(x→−∞) ((2+(√(3+3x^(−1) )))/(1−x^(−1) ))  = 2+(√3)
$$\:\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}}}{\mathrm{x}−\mathrm{1}}\right)=? \\ $$$$\mathrm{Sol}\::\: \\ $$$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}−\sqrt{\mathrm{3x}^{\mathrm{2}} +\mathrm{3x}}}{\mathrm{x}−\mathrm{1}}\right) \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\frac{\mathrm{2x}−\mid\mathrm{x}\mid\sqrt{\mathrm{3}+\frac{\mathrm{3}}{\mathrm{x}}}}{\mathrm{x}−\mathrm{1}}\right) \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{x}\left(\mathrm{2}+\sqrt{\mathrm{3}+\mathrm{3x}^{−\mathrm{1}} }\right.}{\mathrm{x}\left(\mathrm{1}−\mathrm{x}^{−\mathrm{1}} \right)} \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{2}+\sqrt{\mathrm{3}+\mathrm{3x}^{−\mathrm{1}} }}{\mathrm{1}−\mathrm{x}^{−\mathrm{1}} } \\ $$$$=\:\mathrm{2}+\sqrt{\mathrm{3}}\: \\ $$
Answered by Olaf_Thorendsen last updated on 26/Jun/21
f(x) = ((2x−(√(3x^2 +3x)))/(x−1))  f(x) = ((4x^2 −(3x^2 +3x))/((x−1)(2x+(√(3x^2 +3x)))))  f(x) = ((x^2 −3x)/((x−1)(2x+(√(3x^2 +3x)))))  f(x) = ((x^2 −3x)/((x−1)(2x+∣x∣(√(3+(3/x))))))  lim_(x→−∞) f(x)  = lim_(x→−∞) ((x^2 −3x)/((x−1)(2x−x(√(3+(3/x))))))  = lim_(x→−∞) ((x−3)/((x−1)(2−(√(3+(3/x))))))  = (1/(2−(√3))) = 2+(√3)
$${f}\left({x}\right)\:=\:\frac{\mathrm{2}{x}−\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}}}{{x}−\mathrm{1}} \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{4}{x}^{\mathrm{2}} −\left(\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}\right)}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}}\right)} \\ $$$${f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{x}}\right)} \\ $$$${f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}+\mid{x}\mid\sqrt{\mathrm{3}+\frac{\mathrm{3}}{{x}}}\right)} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}{f}\left({x}\right) \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}^{\mathrm{2}} −\mathrm{3}{x}}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}−{x}\sqrt{\mathrm{3}+\frac{\mathrm{3}}{{x}}}\right)} \\ $$$$=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{{x}−\mathrm{3}}{\left({x}−\mathrm{1}\right)\left(\mathrm{2}−\sqrt{\mathrm{3}+\frac{\mathrm{3}}{{x}}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:\mathrm{2}+\sqrt{\mathrm{3}} \\ $$
Answered by mathmax by abdo last updated on 26/Jun/21
=lim_(x→−∞) (((2x)/(x−1))−((∣x∣(√(3+(3/x))))/(x−1)))  =2+lim_(x→−∞)    (x/(x−1))(√(3+(3/x)))=2+(√3)
$$=\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \left(\frac{\mathrm{2x}}{\mathrm{x}−\mathrm{1}}−\frac{\mid\mathrm{x}\mid\sqrt{\mathrm{3}+\frac{\mathrm{3}}{\mathrm{x}}}}{\mathrm{x}−\mathrm{1}}\right) \\ $$$$=\mathrm{2}+\mathrm{lim}_{\mathrm{x}\rightarrow−\infty} \:\:\:\frac{\mathrm{x}}{\mathrm{x}−\mathrm{1}}\sqrt{\mathrm{3}+\frac{\mathrm{3}}{\mathrm{x}}}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$

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