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Question Number 131370 by EDWIN88 last updated on 10/Feb/21

lim_(x→−∞) (√(4x^4 +6x^2 )) +x(√(4x^2 +2)) =?

$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\sqrt{\mathrm{4}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{2}}\:=? \\ $$

Answered by JDamian last updated on 04/Feb/21

∞

$$\infty \\ $$

Answered by EDWIN88 last updated on 10/Feb/21

lim_(x→−∞) (√(x^2 (4x^2 +6))) +x(√(x^2 (4+(2/x^2 )))) = lim_(x→−∞) −x(√(4x^2 +6)) −x^2 (√(4+(2/x^2 ))) = lim_(x→−∞) x^2 (√(4+(6/x^2 )))−x^2 (√(4+(2/x^2 ))) = let (1/x^2 ) = h ∧ h→0 lim_(h→0) (((√(4+6h^2 ))−(√(4+2h^2 )))/h^2 ) = 2×lim_(h→0) (((√(1+(3/2)h^2 ))−(√(1+(h^2 /2))))/h) = 2×lim_(h→0) (((1+((3h^2 )/4))−(1+(h^2 /4)))/h^2 )= 1

$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{4x}^{\mathrm{2}} +\mathrm{6}\right)}\:+\mathrm{x}\sqrt{\mathrm{x}^{\mathrm{2}} \left(\mathrm{4}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }\right)}\:= \\ $$$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}−\mathrm{x}\sqrt{\mathrm{4x}^{\mathrm{2}} +\mathrm{6}}\:−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }}\:= \\ $$$$\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}x}^{\mathrm{2}} \sqrt{\mathrm{4}+\frac{\mathrm{6}}{\mathrm{x}^{\mathrm{2}} }}−\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{2}} }}\:= \\ $$$$\:\mathrm{let}\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{h}\:\wedge\:\mathrm{h}\rightarrow\mathrm{0} \\ $$$$\:\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{4}+\mathrm{6h}^{\mathrm{2}} }−\sqrt{\mathrm{4}+\mathrm{2h}^{\mathrm{2}} }}{\mathrm{h}^{\mathrm{2}} }\:=\:\mathrm{2}×\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}}\mathrm{h}^{\mathrm{2}} }−\sqrt{\mathrm{1}+\frac{\mathrm{h}^{\mathrm{2}} }{\mathrm{2}}}}{\mathrm{h}} \\ $$$$=\:\mathrm{2}×\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{3h}^{\mathrm{2}} }{\mathrm{4}}\right)−\left(\mathrm{1}+\frac{\mathrm{h}^{\mathrm{2}} }{\mathrm{4}}\right)}{\mathrm{h}^{\mathrm{2}} }=\:\mathrm{1}\: \\ $$

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