lim-x-1-4x-5-x-x-2-

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