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

Question Number 165699 by lapache last updated on 06/Feb/22

l i \underset{x \to 2}{m} \frac{x - \sqrt{x + 2}}{\sqrt{4 x + 1} - 3} = \dots . ? ? ?

Commented by cortano1 last updated on 07/Feb/22

\lim_{x \to 2} \frac{x - \sqrt{x + 2}}{\sqrt{4 x + 1} - 3} \times \frac{\sqrt{4 x + 1} + 3}{\sqrt{4 x + 1} + 3} \times \frac{x + \sqrt{x + 2}}{x + \sqrt{x + 2}}

= \lim_{x \to 2} \frac{\sqrt{4 x + 1} + 3}{x + \sqrt{x + 2}} \times \lim_{x \to 2} \frac{x^{2} - x - 2}{4 x - 8}

= \frac{6}{4} \times \lim_{x \to 2} \frac{(x - 2) (x + 1)}{4 (x - 2)} = \frac{3}{2} \times \frac{3}{4}

= \frac{9}{8}

Answered by nurtani last updated on 06/Feb/22

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

= \frac{1 - \frac{1}{2 \sqrt{2 + 2}}}{\frac{4}{2 \sqrt{4 (2) + 1}}} = \frac{1 - \frac{1}{2 (2)}}{\frac{4}{2 \sqrt{9}}} = \frac{1 - \frac{1}{4}}{\frac{4}{6}} = \frac{\frac{3}{4}}{\frac{2}{3}} = \frac{9}{8}

Facebook Tweet Pin