lim-x-0-x-2-x-2-2-1-x-

Question Number 141779 by iloveisrael last updated on 23/May/21

\lim_{x \to 0} \frac{x^{2}}{x + 2 - 2 \sqrt{1 + x}} = ?

Answered by Willson last updated on 23/May/21

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

= \lim_{x \to 0} \frac{2 x \sqrt{1 + x}}{\sqrt{1 + x} - 1}

= \lim_{x \to 0} \frac{2 x \sqrt{1 + x} (\sqrt{1 + x} + 1)}{(\sqrt{1 + x} - 1) (\sqrt{1 + x} + 1)}

= \lim_{x \to 0} \frac{2 x \sqrt{1 + x} (\sqrt{1 + x} + 1)}{x}

= \underset{x \to 0}{\lim 2} \sqrt{1 + x} (\sqrt{1 + x} + 1)

= 4

Answered by cherokeesay last updated on 23/May/21

= \lim_{x \to 0} \frac{x^{2} (x + 2 + 2 \sqrt{1 + x})}{{(x + 2)}^{2} - 4 (1 + x)} =

\lim_{x \to 0} \frac{x^{2} (x + 2 + 2 \sqrt{1 + x})}{x^{2} + 4 - 4 x - 4 - 4 x} = 2 + 2 = 4

Answered by iloveisrael last updated on 23/May/21

Answered by mathmax by abdo last updated on 23/May/21

letf (x) = \frac{x^{2}}{x + 2 - 2 \sqrt{1 + x}} \Rightarrow \lim_{x \to 0} f (x)

= \lim_{x \to 0} \frac{x^{2} (x + 2 + 2 \sqrt{1 + x})}{{(x + 2)}^{2} - 4 (1 + x)} = \lim_{x \to 0} \frac{x^{2} (x + 2 + 2 \sqrt{1 + x})}{x^{2} + 4 x + 4 - 4 - 4 x}

= \lim_{x \to 0} x + 2 + 2 \sqrt{1 + x} = 2 + 2 = 4