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Question Number 169013 by 0731619 last updated on 23/Apr/22

Commented by cortano1 last updated on 23/Apr/22

$$=\sqrt{\underset{x\to \mathrm{\infty }}{\lim }\frac{x}{x+\sqrt{x+\sqrt{x}}}}$$$$=\sqrt{\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}}$$$$=\sqrt{\frac{1}{1}}=1$$

Answered by alephzero last updated on 23/Apr/22

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{\frac{x}{x+\sqrt{x+\sqrt{x}}}}=$$$$\sqrt{\underset{x\to \mathrm{\infty }}{\lim }\frac{x}{x+\sqrt{x+\sqrt{x}}}}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{x}{x+\sqrt{x+\sqrt{x}}}=$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{1+{\left(\sqrt{x+\sqrt{x}}\right)}^{\prime }}$$$$\frac{d\left(\sqrt{x+\sqrt{x}}\right)}{dx}=$$$$\frac{d\left(\sqrt{x+\sqrt{x}}\right)}{d(x+\sqrt{x})}\cdot \frac{d(x+\sqrt{x})}{dx}=$$$$=\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}})$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}})}$$$$\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}})\to 0asx\to \mathrm{\infty }$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{1}{1+\frac{1}{2\sqrt{x+\sqrt{x}}}(1+\frac{1}{2\sqrt{x}})}=1$$$$\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\frac{\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=1$$

Answered by qaz last updated on 23/Apr/22

$$\underset{\mathrm{x}\to +\mathrm{\infty }}{\lim }\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}=\underset{\mathrm{x}\to +\mathrm{\infty }}{\lim }\sqrt{\frac{1}{1+\frac{1}{\mathrm{x}}\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}=1$$$$\underset{\mathrm{x}\to -\mathrm{\infty }}{\lim }\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}\mathrm{not}\mathrm{exsist}.$$$$\underset{\mathrm{x}\to +\mathrm{\infty }}{\lim }\ne \underset{\mathrm{x}\to -\mathrm{\infty }}{\lim }$$$$\Rightarrow \underset{\mathrm{x}\to \mathrm{\infty }}{\lim }\frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}}}}}\mathrm{not}\mathrm{exsist}.$$

Commented by JDamian last updated on 23/Apr/22

really, qaz? I hope you are not a math teacher

Commented by qaz last updated on 23/Apr/22

$$\mathrm{that}\mathrm{infinity}\mathrm{symbol}\mathrm{is}\mathrm{ambiguous}.\mathrm{it}\mathrm{can}\mathrm{be}\mathrm{positive}\mathrm{infinity}\mathrm{or}$$$$\mathrm{negative}\mathrm{infinity}.$$

Commented by JDamian last updated on 23/Apr/22

so are ambiguous the symbols 1, 2, 3, ...

Commented by JDamian last updated on 23/Apr/22

$$Accordingtoyourreasoning,$$$$\underset{x\to \mathrm{\infty }}{\lim }{e}^{-x}doesnotexistbecause$$$$\underset{x\to -\mathrm{\infty }}{\lim }{e}^{-x}\ne \underset{x\to +\mathrm{\infty }}{\lim }{e}^{-x},{doesn}^{\prime }tit?$$

Commented by qaz last updated on 23/Apr/22

$$\mathrm{yes},\mathrm{for}\mathrm{function},\mathrm{x}\to \mathrm{\infty },{}^{\prime }{\mathrm{x}}^{\prime }\mathrm{can}\mathrm{be}\mathrm{positive}\mathrm{or}\mathrm{negative}.$$$$\mathrm{for}\mathrm{sequence},\mathrm{n}\to \mathrm{\infty },\mathrm{here}{}^{\prime }{\mathrm{n}}^{\prime }\mathrm{is}\mathrm{always}\mathrm{positive}.$$

Commented by JDamian last updated on 23/Apr/22

It is the first time I read that

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