Lim-x-16x-2-2x-1-4x-5-

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

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

Commented by MJS_new last updated on 13/Mar/23

a > 0 \Rightarrow \lim_{x \to \infty} \sqrt{{a x}^{2} + b x + c} - (\sqrt{a} x + d) = \frac{b}{2 \sqrt{a}} - d

Commented by cortano12 last updated on 13/Mar/23

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

yes, thank you

Answered by cortano12 last updated on 10/Mar/23

\lim_{x \to \infty} \sqrt{{16 x}^{2} - 2 x - 1} - 4 x - 5

= \lim_{x \to \infty} 4 x [\sqrt{1 - \frac{1}{8 x} - \frac{1}{{16 x}^{2}}} - 1 - \frac{5}{4 x}]

[\frac{1}{4 x} = h; h \to 0]

= \lim_{h \to 0} \frac{1}{h} [\sqrt{1 - \frac{1}{2} h - h^{2}} - 1 - 5 h]

= - 5 + \lim_{h \to 0} \frac{\sqrt{1 - \frac{1}{2} h - h^{2}} - 1}{h}

= - 5 + \frac{1}{2} \lim_{h \to 0} \frac{h (- \frac{1}{2} - h)}{h}

= - \frac{21}{4}