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Question Number 120720 by bramlexs22 last updated on 02/Nov/20

Given lim_(x→∞) (√(x+2(√x)+3)) −(√x)+b = 4 find the value of b^2 +1

$$\:\mathrm{Given}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{\mathrm{x}+\mathrm{2}\sqrt{\mathrm{x}}+\mathrm{3}}\:−\sqrt{\mathrm{x}}+\mathrm{b}\:=\:\mathrm{4} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{b}^{\mathrm{2}} +\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 02/Nov/20

(√x)((√(1+(2/( (√x)))+(3/x)))−1+(b/( (√x))))=4 lim_(x→∞) (√x)(1+(1/( (√x)))+(3/(2x))−1+(b/( (√x))))=4 (√x)(((1+b)/( (√x))))=4 b=3 b^2 +1=10

$$\sqrt{{x}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\:\sqrt{{x}}}+\frac{\mathrm{3}}{{x}}}−\mathrm{1}+\frac{{b}}{\:\sqrt{{x}}}\right)=\mathrm{4} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{x}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{{x}}}+\frac{\mathrm{3}}{\mathrm{2}{x}}−\mathrm{1}+\frac{{b}}{\:\sqrt{{x}}}\right)=\mathrm{4} \\ $$$$\sqrt{{x}}\left(\frac{\mathrm{1}+{b}}{\:\sqrt{{x}}}\right)=\mathrm{4} \\ $$$${b}=\mathrm{3} \\ $$$${b}^{\mathrm{2}} +\mathrm{1}=\mathrm{10} \\ $$

Answered by john santu last updated on 02/Nov/20

lim_(x→∞) (√(x+2(√x)+3)) −((√x)−b)=4 lim_(x→∞) (√(x+2(√x) +3))−(√(((√x)−b)^2 )) = 4 lim_(x→∞) (√(x+2(√x)+3)) −(√(x−2b(√x)+b^2 )) = 4 ((2−(−2b))/(2.1)) = 4 ⇒2+2b=8 ⇒b=3 ∧b^2 +1=10

$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}+\mathrm{2}\sqrt{{x}}+\mathrm{3}}\:−\left(\sqrt{{x}}−{b}\right)=\mathrm{4} \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{x}+\mathrm{2}\sqrt{{x}}\:+\mathrm{3}}−\sqrt{\left(\sqrt{{x}}−{b}\right)^{\mathrm{2}} }\:=\:\mathrm{4} \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt{{x}+\mathrm{2}\sqrt{{x}}+\mathrm{3}}\:−\sqrt{{x}−\mathrm{2}{b}\sqrt{{x}}+{b}^{\mathrm{2}} }\:=\:\mathrm{4} \\ $$$$\:\frac{\mathrm{2}−\left(−\mathrm{2}{b}\right)}{\mathrm{2}.\mathrm{1}}\:=\:\mathrm{4}\:\Rightarrow\mathrm{2}+\mathrm{2}{b}=\mathrm{8}\:\Rightarrow{b}=\mathrm{3}\:\wedge{b}^{\mathrm{2}} +\mathrm{1}=\mathrm{10} \\ $$

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