lim-x-0-x-x-x-x-

Question Number 12732 by Joel577 last updated on 30/Apr/17

\lim_{x \to 0} \frac{\sqrt{x} - x}{\sqrt{x} + x}

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/Apr/17

= 1

Commented by Joel577 last updated on 30/Apr/17

t h a n k y o u

Answered by ridwan balatif last updated on 30/Apr/17

\frac{\sqrt{x} - x}{\sqrt{x} + x} \times \frac{\sqrt{x} + x}{\sqrt{x} + x}

\frac{x - x^{2}}{x + 2 x \sqrt{x} + x^{2}}

\frac{x (1 - x)}{x (1 + 2 \sqrt{x} + x)}

\frac{1 - x}{1 + 2 \sqrt{x} + x}

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

Commented by Joel577 last updated on 30/Apr/17

t h a n k y o u v e r y m u c h

Answered by ajfour last updated on 30/Apr/17

R . H . L . = \lim_{x \to 0} \frac{\sqrt{x} (1 - \sqrt{x})}{\sqrt{x} (1 + \sqrt{x})}

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

= 1

Commented by Joel577 last updated on 30/Apr/17

t h a n k y o u v e r y m u c h

Answered by malwaan last updated on 30/Apr/17

put x = y^{2}

x \to 0 \Rightarrow y \to 0

\lim_{y \to 0} \frac{y - y^{2}}{y + y^{2}} = \lim_{y \to 0} \frac{y (1 - y)}{y (1 + y)}

= \lim_{y \to 0} \frac{1 - y}{1 + y} = \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1

Commented by Joel577 last updated on 01/May/17

t h a n k y o u v e r y m u c h

Facebook Tweet Pin