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Question Number 197407 by mathlove last updated on 16/Sep/23

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

Answered by Rasheed.Sindhi last updated on 16/Sep/23

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

Commented by mathlove last updated on 16/Sep/23

$$thanks$$

Answered by MM42 last updated on 16/Sep/23

$${lim}_{x\to \mathrm{\infty }}\frac{\sqrt{x^2(1+\frac{1}{x^2})}}{x+1}={lim}_{x\to \mathrm{\infty }}\frac{\mid x\mid \sqrt{1+\frac{1}{x^2}}}{x(1+\frac{1}{x})}$$$$={lim}_{x\to \mathrm{\infty }}\frac{\mid x\mid }{x}=\{\begin{array}{l}1;x\to +\mathrm{\infty }\\ -1;x\to -\mathrm{\infty }\end{array}$$$$\Rightarrow lim:notexist$$

Commented by MathematicalUser2357 last updated on 17/Sep/23

$$\mathrm{So}\underset{x\to \mathrm{\infty }}{\lim }\frac{\mid x\mid }{x}\ne 1$$

Commented by MM42 last updated on 18/Sep/23

$${lim}_{x\to +\mathrm{\infty }}\frac{x}{x}=1\mathrm{\&}{lim}_{x\to -\mathrm{\infty }}\frac{-x}{x}=-1$$

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