Lim x-->0 [√(2+x) - √(2-x)] / x.... Question from @Galladeaviero

December 2018 2 92 Report

Lim x-->0 [√(2+x) - √(2-x)] / x

HenryThomas Lim x-->0 √2+x - √2-x / x
lim x-->0 √2+x - √2-x / x (√2+x + √2-x /√2+x + √2-x )
lim x-->0 2+x-2+x / x(√2+x + √2-x)
lim x-->0 2x / x(√2+x + √2-x)
lim x-->0 2 / √2+x + √2-x
2 / √2+0 + √2-0
2/ √2 + √2
2/ 2√2 = 1/2 √2

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whongaliem $\lim_{x \to \.0} \frac{ \sqrt{2 + x} - \sqrt{2 - x} }{x}$ =
$\lim_{x \to \.0} \frac{ \sqrt{2 + x} - \sqrt{2 - x} }{x} X \frac{ \sqrt{2 + x} + \sqrt{2 - x} }{ \sqrt{2 + x} + \sqrt{2 - x} }$ =
$\lim_{x \to \.0} \frac{2 + x - 2 + x}{x ( \sqrt{2 + x} + \sqrt{2 - x} }$ =
$\lim_{x \to \.0} \frac{2x}{x ( \sqrt{2 +x} + \sqrt{2 - x} }$ =
$\lim_{x \to \.0} \frac{2}{ \sqrt{2 + x} + \sqrt{2 - x} }$ =
$\frac{2}{ \sqrt{2 + 0} + \sqrt{2 + 0} }$ =
$\frac{2}{2 \sqrt{2} }$ = $\frac{1}{2} \sqrt{2}$

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