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Question Number 121172 by benjo_mathlover last updated on 05/Nov/20

lim_(x→∞) (((√(x^2 +1))−x+1)/(x+1)) ?

$$\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{x}+\mathrm{1}}{\mathrm{x}+\mathrm{1}}\:? \\ $$$$ \\ $$

Answered by TANMAY PANACEA last updated on 05/Nov/20

lim_(x→∞) ((x(√(1+(1/x^2 ))) −x+1)/(x+1)) lim_(x→∞) (((√(1+(1/x^2 ))) −1+(1/x))/(1+(1/x)))=(((√(1+0)) −1)/1)=0

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:−{x}+\mathrm{1}}{{x}+\mathrm{1}}\: \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}\:−\mathrm{1}+\frac{\mathrm{1}}{{x}}}{\mathrm{1}+\frac{\mathrm{1}}{{x}}}=\frac{\sqrt{\mathrm{1}+\mathrm{0}}\:−\mathrm{1}}{\mathrm{1}}=\mathrm{0} \\ $$

Answered by Bird last updated on 05/Nov/20

f(x)=(((√(x^2 +1))−x+1)/(x+1)) ⇒ f(x)=((x(√(1+(1/x^2 )))−x+1)/(x+1)) ∼((x(1+(1/(2x^2 )))−x+1)/(x+1))=(((1/(2x))+1)/(x+1)) ⇒ lim_(x→+∞) f(x)=0

$${f}\left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}−{x}+\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{{x}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−{x}+\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\sim\frac{{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }\right)−{x}+\mathrm{1}}{{x}+\mathrm{1}}=\frac{\frac{\mathrm{1}}{\mathrm{2}{x}}+\mathrm{1}}{{x}+\mathrm{1}}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\mathrm{0} \\ $$

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