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Question Number 188984 by normans last updated on 10/Mar/23

$$$$$${\mathbf{Lim}}_{\mathit{x}\to \sim }\sqrt{16{\mathit{x}}^2-2\mathit{x}-1}-4\mathit{x}-5=??$$$$$$

Commented by MJS_new last updated on 13/Mar/23

$$a>0\Rightarrow \underset{x\to \mathrm{\infty }}{\lim }\sqrt{{ax}^2+bx+c}-(\sqrt{a}x+d)=\frac{b}{2\sqrt{a}}-d$$

Commented by cortano12 last updated on 13/Mar/23

$$\mathrm{it}\mathrm{should}\mathrm{be}\mathrm{L}=\frac{\mathrm{b}}{2\sqrt{\mathrm{a}}}-\mathrm{d}$$$$$$

Commented by cortano12 last updated on 13/Mar/23

$$=\frac{-2}{2\sqrt{16}}-5=-\frac{1}{4}-5=-\frac{21}{4}$$

Commented by MJS_new last updated on 13/Mar/23

$$\mathrm{yes},\mathrm{thank}\mathrm{you}$$

Answered by cortano12 last updated on 10/Mar/23

$$\underset{x\to \mathrm{\infty }}{\lim }\sqrt{{16\mathrm{x}}^2-2\mathrm{x}-1}-4\mathrm{x}-5$$$$=\underset{x\to \mathrm{\infty }}{\lim }4\mathrm{x}[\sqrt{1-\frac{1}{8\mathrm{x}}-\frac{1}{{16\mathrm{x}}^2}}-1-\frac{5}{4\mathrm{x}}]$$$$[\frac{1}{4\mathrm{x}}=\mathrm{h};\mathrm{h}\to 0]$$$$=\underset{\mathrm{h}\to 0}{\lim }\frac{1}{\mathrm{h}}[\sqrt{1-\frac{1}{2}\mathrm{h}-{\mathrm{h}}^2}-1-5\mathrm{h}]$$$$=-5+\underset{\mathrm{h}\to 0}{\lim }\frac{\sqrt{1-\frac{1}{2}\mathrm{h}-{\mathrm{h}}^2}-1}{\mathrm{h}}$$$$=-5+\frac{1}{2}\underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{h}(-\frac{1}{2}-\mathrm{h})}{\mathrm{h}}$$$$=-\frac{21}{4}$$

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