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Question Number 164266 by naka3546 last updated on 15/Jan/22

$$\underset{x\to \mathrm{\infty }}{\lim }(x-2)-\sqrt{x^2+2x-5}=?$$

Answered by ajfour last updated on 15/Jan/22

$$x=\frac{1}{h}\to \mathrm{\infty }$$$$L=\underset{h\to 0}{\lim }\frac{1-2h-{(1+2h-5h^2)}^{1/2}}{h}$$$$=\underset{h\to 0}{\lim }\frac{1-2h-(1+\frac{2h-5h^2}{2})}{h}$$$$=\underset{h\to 0}{\lim }\frac{-4h-2h+5h^2}{2h}=-3$$

Commented by naka3546 last updated on 15/Jan/22

$$Thankyou,sir.$$

Answered by cortano1 last updated on 16/Jan/22

$$\underset{x\to \mathrm{\infty }}{\lim }(x-2)-(x+\frac{2}{2.1})=$$$$\underset{x\to \mathrm{\infty }}{\lim }x-2-x-1=-3$$

Commented by naka3546 last updated on 16/Jan/22

$$thankyou,sir.$$

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