lim_(x→2^− ) ((2−2cos (√(x−2)))/((x−2)^2 )) =?


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Question Number 161586 by blackmamba last updated on 20/Dec/21

$\lim_{x \to 2^{-}} \frac{2 - 2 \cos \sqrt{x - 2}}{{(x - 2)}^{2}} = ?$
Commented by cortano last updated on 20/Dec/21

Commented by mathmax by abdo last updated on 20/Dec/21

$sir cortano you take another limit loook the Q . . .$
	Answered by TheSupreme last updated on 20/Dec/21

	$t = x - 2$ $l i m \frac{2 - 2 c o s (\sqrt{t})}{t^{2}} = 2 \frac{1 - c o s (\sqrt{t})}{t^{2}} = 2 \frac{1 - (1 - \frac{t}{2})}{t^{2}} = \infty$
	Commented by cortano last updated on 20/Dec/21

	$t h e q u e s t i o n$ $\lim_{x \to 2^{-}} \frac{x - 2 \cos \sqrt{2 - x}}{{(2 - x)}^{2}}$ $\sqrt{2 - x} d e f i n e d o n 2 - x ⩾ 0$ $\Rightarrow x ⩽ 2$
	Commented by cortano last updated on 20/Dec/21

	$w h y n o t v a l i d ?$
	Commented by cortano last updated on 20/Dec/21

	$A n s w e r - \frac{1}{12}$
	Commented by mr W last updated on 20/Dec/21

	$- \frac{1}{12} i s w r o n g .$ $\lim_{x \to 2^{+}} (. . .) = + \infty$ $x < 2 i s n o t v a l i d .$
	Commented by mnjuly1970 last updated on 20/Dec/21

	$D_{f} = (\infty, 2]$
	Commented by mr W last updated on 20/Dec/21

	$c o r t a n o s i r :$ $y o u c h a n g e d t h e q u e s t i o n b y y o u r s e l f .$ $t h e q u e s t i o n i s$ $\frac{2 - 2 \cos \sqrt{x - 2}}{{(x - 2)}^{2}}, n o t \frac{x - 2 \cos \sqrt{2 - x}}{{(2 - x)}^{2}}!$
	Answered by mathmax by abdo last updated on 20/Dec/21

	$f (x) = \frac{2 - 2 \cos \sqrt{x - 2}}{{(x - 2)}^{2}} we do the changement \sqrt{x - 2} = t \Rightarrow$ $f (x) = \frac{2 - 2 cost}{t^{4}} (x \to 2 \Rightarrow t \to 0)$ $cost \sim 1 - \frac{t^{2}}{2} \Rightarrow f (x) \sim \frac{2 - 2 + t^{2}}{t^{4}} = \frac{1}{t^{2}} \to + \infty$
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