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bemath-lim-x-x-4x-2-1-x-2-2x-1-




Question Number 111351 by bemath last updated on 03/Sep/20
   (√(bemath))  lim_(x→−∞)  ((x−(√(4x^2 +1)))/( (√(x^2 +2x+1)))) ?
$$\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{x}−\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{1}}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}}}\:? \\ $$
Answered by ajfour last updated on 03/Sep/20
     =lim_(x→−∞) (((x/(∣x∣))−(√(4+(1/x^2 ))))/( (√(1+((2x)/(∣x∣^2 ))+(1/x^2 ))))) = ((−1−2)/1)       =−3 .
$$\:\:\:\:\:=\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\frac{{x}}{\mid{x}\mid}−\sqrt{\mathrm{4}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{2}{x}}{\mid{x}\mid^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}}\:=\:\frac{−\mathrm{1}−\mathrm{2}}{\mathrm{1}} \\ $$$$\:\:\:\:\:=−\mathrm{3}\:. \\ $$
Answered by bobhans last updated on 03/Sep/20
 lim_(x→−∞)  ((x−(√(x^2 (4+(1/x^2 )))))/( (√(x^2 (1+(2/x)+(1/x^2 )))))) = lim_(x→−∞) ((x+x(√(4+(1/x^2 ))))/(−x(√(1+(2/x)+(1/x^2 )))))  lim_(x→−∞) ((1+(√(4+(1/x^2 ))))/(−(√(1+(2/x)+(1/x^2 ))))) = ((1+2)/(−1)) = −3
$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\frac{\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{4}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{x}+\mathrm{x}\sqrt{\mathrm{4}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}{−\mathrm{x}\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}} \\ $$$$\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\frac{\mathrm{1}+\sqrt{\mathrm{4}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}{−\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}}\:=\:\frac{\mathrm{1}+\mathrm{2}}{−\mathrm{1}}\:=\:−\mathrm{3} \\ $$

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