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

Question Number 155316 by ZiYangLee last updated on 28/Sep/21

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}−{x}\right)=? \\ $$

Answered by puissant last updated on 28/Sep/21

=lim_(x→∞) ((3x)/( x+(√(x^2 +3x)))) = lim_(x→∞) (x/x)×(3/(1+(√(1+(3/x))))) =1×lim_(x→∞) (3/(1+(√(1+(3/x)))))= (3/2)..

$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}{x}}{\:{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{{x}}{{x}}×\frac{\mathrm{3}}{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}}} \\ $$$$=\mathrm{1}×\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}}{\mathrm{1}+\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}}}=\:\frac{\mathrm{3}}{\mathrm{2}}.. \\ $$

Answered by amin96 last updated on 28/Sep/21

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

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}−{x}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}\right)}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}}\right)=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}}= \\ $$$$=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{3}}{\:\sqrt{\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{{x}}}+\mathrm{1}}=\frac{\underset{{x}\rightarrow\infty} {\mathrm{lim}3}}{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{3}}{\mathrm{1}+\mathrm{1}}=\mathrm{1},\mathrm{5} \\ $$

Answered by yeti123 last updated on 28/Sep/21

using formula: lim_(x→∞) (√(ax^2 + bx + c)) − (√(ax^2 + px + q)) = ((b−p)/(2(√a))) in this case lim_(x→∞) (√(x^2 + 3x)) − x = lim_(x→∞) (√(x^2 + 3x)) − (√x^2 ) = ((3−0)/(2(√1))) = (3/2)

$$\mathrm{using}\:\mathrm{formula}: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{ax}^{\mathrm{2}} \:+\:{bx}\:+\:{c}}\:−\:\sqrt{{ax}^{\mathrm{2}} \:+\:{px}\:+\:{q}}\:=\:\frac{{b}−{p}}{\mathrm{2}\sqrt{{a}}} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}\:−\:{x}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{3}{x}}\:−\:\sqrt{{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{3}−\mathrm{0}}{\mathrm{2}\sqrt{\mathrm{1}}}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by physicstutes last updated on 28/Sep/21

lim_(x→∞) [((((√(x^2 +3x))−x)((√(x^2 +3x))+x))/(((√(x^2 +3x))+x)))] = lim_(x→∞) [((x^2 +3x−x^2 )/( (√(x^2 +3x))+x))] = lim_(x→∞) (((3x)/(x((√(1+(3/x)))+1)))=(3/(1+1)) =(3/2)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}−{x}\right)\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}\right)}{\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}\right)}\right] \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{x}^{\mathrm{2}} +\mathrm{3}{x}−{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}}+{x}}\right] \\ $$$$=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{3}{x}}{{x}\left(\sqrt{\mathrm{1}+\frac{\mathrm{3}}{{x}}}+\mathrm{1}\right.}\right)=\frac{\mathrm{3}}{\mathrm{1}+\mathrm{1}}\:=\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\: \\ $$

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