lim_(x→1^+ ) ((4x+5)/(x−x^2 ))=?


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Question Number 195092 by mathlove last updated on 24/Jul/23

$\lim_{x \to 1^{+}} \frac{4 x + 5}{x - x^{2}} = ?$
	Answered by tri26112004 last updated on 24/Jul/23

	$W e h a v e :$ $∙ \lim_{x \to 1^{+}} 4 x + 5 = 4.1 + 5 = 9 > 0$ $∙ \lim_{x \to 1^{+}} x - x^{2} = 0 (\forall x > 1)$ $∙ g (x) = x - x^{2} - > a = - 1 < 0$ $\Rightarrow \lim_{x \to 1^{+}} \frac{4 x + 5}{x - x^{2}} = - \infty$
	Commented by mathlove last updated on 24/Jul/23

	$w a y a = - 1$ $a s s x \to 1^{+} g (0.9) = 0, 9 - {(0.9)}^{2} = 0.09$ $v a l u e i s t h e +$ $w a y t h e l i m i t s y m p t o m i s$ $- \infty$
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