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Question Number 7243 by Sh505146@gmail.com last updated on 18/Aug/16

lim_(x→∞) ((√(x^2 +x+1)))−((√(x^2 +1))) <

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\left.{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}−\left(\sqrt{\left.{x}^{\mathrm{2}} +\mathrm{1}\right)}\right.\right. \\ $$ $$< \\ $$

Commented byRasheed Soomro last updated on 18/Aug/16

lim_(x→∞) ((√(x^2 +x+1))−(√(x^2 +1))) ((√(x^2 +x+1))−(√(x^2 +1)))= ((((√(x^2 +x+1))−(√(x^2 +1)))((√(x^2 +x+1))+(√(x^2 +1))))/((√(x^2 +x+1))+(√(x^2 +1)))) =((x^2 +x+1−x^2 −1)/((√(x^2 +x+1))+(√(x^2 +1)))) =(x/((√(x^2 +x+1))+(√(x^2 +1)))) Let x=(1/y) (x/((√(x^2 +x+1))+(√(x^2 +1))))=((1/y)/((√(((1/y))^2 +(1/y)+1))+(√(((1/y))^2 +1)))) =((1/y)/((√((1/y^2 )+(1/y)+1))+(√((1/y^2 )+1))))=((1/y)/(((√(1+y+y^2 ))+(√(1+y^2 )))/y)) =(1/((√(1+y+y^2 ))+(√(1+y^2 )))) Now as x=1/y , So x→∞ ⇒y→0 Hence lim_(x→∞) ((√(x^2 +x+1))−(√(x^2 +1)))=lim_(y→0) (1/((√(1+y+y^2 ))+(√(1+y^2 )))) =(1/((√(1+0+0^2 ))+(√(1+0^2 ))))=(1/2)

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$ $$\:\:\:\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)=\:\:\:\frac{\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$ $$\:\:\:\:\:\:=\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$ $$\:\:\:\:\:\:=\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$ $${Let}\:{x}=\frac{\mathrm{1}}{{y}} \\ $$ $$\:\:\:\:\:\:\:\:\:\frac{{x}}{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}=\frac{\frac{\mathrm{1}}{{y}}}{\sqrt{\left(\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{y}}+\mathrm{1}}+\sqrt{\left(\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} +\mathrm{1}}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{1}}{{y}}}{\sqrt{\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}}+\mathrm{1}}+\sqrt{\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\mathrm{1}}}=\frac{\frac{\mathrm{1}}{{y}}}{\frac{\sqrt{\mathrm{1}+{y}+{y}^{\mathrm{2}} }+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{y}+{y}^{\mathrm{2}} }+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }} \\ $$ $${Now}\:{as}\:{x}=\mathrm{1}/{y}\:,\:{So}\:{x}\rightarrow\infty\:\Rightarrow{y}\rightarrow\mathrm{0} \\ $$ $${Hence}\: \\ $$ $$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)=\underset{{y}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{y}+{y}^{\mathrm{2}} }+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{0}+\mathrm{0}^{\mathrm{2}} }+\sqrt{\mathrm{1}+\mathrm{0}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented bypeter james last updated on 19/Aug/16

welldone..

$${welldone}.. \\ $$

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