find δ>0 such that ∣f(x)+1∣<0.01 when 0<∣x−2∣<δ,where f(x)=((x^2 −5x+6)/(x−2)),hence use ε_δ definition to show that ((lim)/(xtends 2))f(x)=−1


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Question Number 25814 by rita1608 last updated on 15/Dec/17

$f i n d δ > 0 s u c h t h a t ∣ f (x) + 1 ∣< 0.01$ $w h e n 0 <∣ x - 2 ∣< δ, w h e r e$ $f (x) = \frac{x^{2} - 5 x + 6}{x - 2}, h e n c e u s e ε_δ$ $d e f i n i t i o n t o s h o w t h a t$ $\frac{l i m}{x t e n d s 2} f (x) = - 1$
	Answered by ajfour last updated on 15/Dec/17

	$∣ f (x) + 1 ∣=∣ \frac{(x - 2) (x - 3)}{(x - 2)} + 1 ∣< 0.01$ $\Rightarrow f o r x \neq 2, ∣ x - 3 + 1 ∣< 0.01$ $o r ∣ x - 2 ∣< 0.01$ $s o h e r e δ = ϵ = 0.01$ $\Rightarrow \lim_{x \to 2} f (x) = - 1 .$
	Commented by kaivan.ahmadi last updated on 15/Dec/17

	$δ ⩽ ϵ and Max (δ) = 0.01$
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