If f(x) = ((x^2 −4)/(x−2)) x→2 α = 3 . Σ = 0.1 find δ Alpha = 3 Epsalum = 0.1 Find delta


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Question Number 5232 by sanusihammed last updated on 02/May/16

$I f f (x) = \frac{x^{2} - 4}{x - 2}$ $x \to 2$ $α = 3 . Σ = 0.1 f i n d δ$ $A l p h a = 3 E p s a l u m = 0.1 F i n d d e l t a$
Commented by FilupSmith last updated on 02/May/16

$What is the definition of δ ?$
	Answered by Yozzii last updated on 02/May/16
	$According to the ε−δ definition of the limit, the limit lim_(x→a) f(x)=α exists if,for any ε>0 there is a δ>0, such that ∣f(x)−α∣<ε whenever 0<∣x−a∣<δ. So using f(x)=((x^2 −4)/(x−2)) and α=4 {not 3 since lim_(x→2) f(x)=lim_(x→2) (((x+2)(x−2))/(x−2))=lim_(x→2) (x+2)=2+2=4≠3.} ⇒∣((x^2 −4)/(x−2))−4∣=∣((x^2 −4−4x+8)/(x−2))∣ =∣((x^2 −4x+4)/(x−2))∣=∣(((x−2)^2 )/(x−2))∣ =∣x−2∣ Since ∣f(x)−4∣<ε and ∣f(x)−4∣=∣x−2∣ ⇒∣x−2∣<ε⇒ let ε=δ. Hence, if ε=0.1, δ=0.1.$
	$A c c o r d i n g t o t h e ϵ - δ d e f i n i t i o n o f$ $t h e l i m i t, t h e l i m i t \lim_{x \to a} f (x) = α$ $e x i s t s i f, f o r a n y ϵ > 0 t h e r e i s a δ > 0,$ $s u c h t h a t$ $∣ f (x) - α ∣< ϵ w h e n e v e r 0 <∣ x - a ∣< δ .$ $S o u s i n g f (x) = \frac{x^{2} - 4}{x - 2} a n d α = 4 {n o t 3 s i n c e \lim_{x \to 2} f (x) = \lim_{x \to 2} \frac{(x + 2) (x - 2)}{x - 2} = \lim_{x \to 2} (x + 2) = 2 + 2 = 4 \neq 3 .}$ $\Rightarrow∣ \frac{x^{2} - 4}{x - 2} - 4 ∣=∣ \frac{x^{2} - 4 - 4 x + 8}{x - 2} ∣$ $=∣ \frac{x^{2} - 4 x + 4}{x - 2} ∣=∣ \frac{{(x - 2)}^{2}}{x - 2} ∣$ $=∣ x - 2 ∣$ $S i n c e ∣ f (x) - 4 ∣< ϵ a n d ∣ f (x) - 4 ∣=∣ x - 2 ∣$ $\Rightarrow∣ x - 2 ∣< ϵ \Rightarrow l e t ϵ = δ . H e n c e, i f$ $ϵ = 0.1, δ = 0.1 .$
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