lim_(x→0) ((∣2x−2∣−∣2x+2∣)/x)


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Question Number 71317 by aliesam last updated on 13/Oct/19

$\underset{x \to 0}{l i m} \frac{∣ 2 x - 2 ∣ - ∣ 2 x + 2 ∣}{x}$
Commented by mathmax by abdo last updated on 13/Oct/19
$=lim_(x→0) (((√((2x−2)^2 ))−(√((2x+2)^2 )))/x) =lim_(x→0) ((((√((2x−2)^2 ))−(√((2x+2)^2 )))((√((2x−2)^2 ))+(√((2x+2)^2 ))))/(x{(√((2x−2)^2 ))+(√((2x+2)^2 ))})) =lim_(x→0) (((2x−2)^2 −(2x+2)^2 )/(x{(√((2x−2)^2 ))+(√((2x+2)^2 ))})) =lim_(x→0) ((4(x^2 −2x+1−x^2 −2x−1})/(x{(√((2x−2)^2 ))+(√((2x+2)^2 ))})) =lim_(x→0) ((−16)/((√((2x−2)^2 ))+(√((2x+2)^2 )))) =((−16)/4) =−4$
$= {l i m}_{x \to 0} \frac{\sqrt{{(2 x - 2)}^{2}} - \sqrt{{(2 x + 2)}^{2}}}{x}$ $= {l i m}_{x \to 0} \frac{(\sqrt{{(2 x - 2)}^{2}} - \sqrt{{(2 x + 2)}^{2}}) (\sqrt{{(2 x - 2)}^{2}} + \sqrt{{(2 x + 2)}^{2}})}{x {\sqrt{{(2 x - 2)}^{2}} + \sqrt{{(2 x + 2)}^{2}}}}$ $= {l i m}_{x \to 0} \frac{{(2 x - 2)}^{2} - {(2 x + 2)}^{2}}{x {\sqrt{{(2 x - 2)}^{2}} + \sqrt{{(2 x + 2)}^{2}}}}$ $= {l i m}_{x \to 0} \frac{4 (x^{2} - 2 x + 1 - x^{2} - 2 x - 1}}{x {\sqrt{{(2 x - 2)}^{2}} + \sqrt{{(2 x + 2)}^{2}}}}$ $= {l i m}_{x \to 0} \frac{- 16}{\sqrt{{(2 x - 2)}^{2}} + \sqrt{{(2 x + 2)}^{2}}} = \frac{- 16}{4} = - 4$
Commented by aliesam last updated on 13/Oct/19

$g o d b l e s s y o u s i r$
Commented by mathmax by abdo last updated on 13/Oct/19

$t h a n k y o u s i r .$
	Answered by ajfour last updated on 13/Oct/19

	$x = h \to 0^{+}$ $L = \lim_{h \to 0} \frac{2 - 2 h - 2 h - 2}{h} = - 4$ $x = - h \to 0^{-}$ $L = \lim_{h \to 0} \frac{2 + 2 h + 2 h - 2}{- h} = - 4$ $h e n c e l i m i t L d o e s e x i s t a n d = - 4 .$
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