bemath-lim-x-x-4x-2-1-x-2-2x-1-

Question Number 111351 by bemath last updated on 03/Sep/20

\sqrt{bemath}

\lim_{x \to - \infty} \frac{x - \sqrt{{4 x}^{2} + 1}}{\sqrt{x^{2} + 2 x + 1}} ?

Answered by ajfour last updated on 03/Sep/20

= \lim_{x \to - \infty} \frac{\frac{x}{∣ x ∣} - \sqrt{4 + \frac{1}{x^{2}}}}{\sqrt{1 + \frac{2 x}{∣ x ∣^{2}} + \frac{1}{x^{2}}}} = \frac{- 1 - 2}{1}

= - 3 .

Answered by bobhans last updated on 03/Sep/20

\lim_{x \to - \infty} \frac{x - \sqrt{x^{2} (4 + \frac{1}{x^{2}})}}{\sqrt{x^{2} (1 + \frac{2}{x} + \frac{1}{x^{2}})}} = \lim_{x \to - \infty} \frac{x + x \sqrt{4 + \frac{1}{x^{2}}}}{- x \sqrt{1 + \frac{2}{x} + \frac{1}{x^{2}}}}

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

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