lim-x-4x-2-16x-1-2x-3-

Question Number 167980 by mathlove last updated on 31/Mar/22

\lim_{x \to \infty} \sqrt{4 x^{2} - 16 x + 1} - 2 x + 3 = ?

Commented by cortano1 last updated on 31/Mar/22

= \lim_{x \to \infty} 2 (x - \frac{16}{8}) - 2 x + 3

= - 4 + 3 = - 1

Commented by mathlove last updated on 31/Mar/22

h o w ? ? ?

Commented by mathlove last updated on 31/Mar/22

i t s a n y w a y ?

Answered by qaz last updated on 05/Apr/22

\lim_{x \to + \infty} \sqrt{{4 x}^{2} - 16 x + 1} - 2 x + 3

= \underset{x \to + \infty}{\lim 2 x} \sqrt{1 - \frac{4}{x} + \frac{1}{{4 x}^{2}}} - 2 x + 3

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

= - 1

\lim_{x \to - \infty} \sqrt{{4 x}^{2} - 16 x + 1} - 2 x + 3

= \lim_{x \to + \infty} \sqrt{{4 x}^{2} + 16 x + 1} + 2 x + 3

= \underset{x \to + \infty}{\lim 2 x} \sqrt{1 + \frac{4}{x} + \frac{1}{{4 x}^{2}}} + 2 x + 3

= \underset{x \to + \infty}{\lim 2 x} (1 + \frac{2}{x} + o (\frac{1}{x})) + 2 x + 3

= + \infty

\Rightarrow \lim_{x \to \infty} \sqrt{{4 x}^{2} - 16 x + 1} - 2 x + 3 not exsist . .

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