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

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{{\mathrm{x}}^2+1}-\mathrm{x}+1}{\mathrm{x}+1}?$$$$$$

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

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{x\sqrt{1+\frac{1}{x^2}}-x+1}{x+1}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{1+\frac{1}{x^2}}-1+\frac{1}{x}}{1+\frac{1}{x}}=\frac{\sqrt{1+0}-1}{1}=0$$

Answered by Bird last updated on 05/Nov/20

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

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