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Question Number 1499 by 112358 last updated on 14/Aug/15

U s e 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 o s h o w t h a t

\lim_{x \to 3} \frac{x}{6 - x} = 1 .

(I^{'} m h o p i n g t o b e t t e r u n d e r s t a n d

t h i s c o n c e p t b y e x a m p l e s o p l e a s e

h e l p m e b y e x p l a n i n g t h e

r e a s o n i n g b e h i n d y o u r s t e p s .)

Commented by 123456 last updated on 14/Aug/15

\forall ϵ > 0, \exists δ, \forall x, 0 <∣ x - 3 ∣< δ, ∣ \frac{x}{6 - x} - 1 ∣< ϵ

Commented by 123456 last updated on 14/Aug/15

\frac{x}{6 - x} - 1 = \frac{x - (6 - x)}{6 - x} = \frac{2 x - 6}{6 - x} = \frac{2 (x - 3)}{6 - x}

Commented by 123456 last updated on 14/Aug/15

\frac{x}{6 - x} = k \Leftrightarrow x = 6 k - k x \Leftrightarrow x = \frac{6 k}{1 + k}

k = 1 + ϵ \Rightarrow x = \frac{6 (1 + ϵ)}{2 + ϵ}

Answered by 123456 last updated on 16/Aug/15

\lim_{x \to 3} \frac{x}{6 - x} = 1 \Rightarrow \exists ϵ > 0, \forall δ, 0 <∣ x - 3 ∣< δ \Rightarrow∣ \frac{x}{6 - x} - 1 ∣< ϵ

lets simplyfy things a little

∣ \frac{x}{6 - x} - 1 ∣=∣ \frac{x - (6 - x)}{6 - x} ∣=∣ \frac{x - 6 + x}{6 - x} ∣=∣ \frac{2 x - 6}{6 - x} ∣

= \frac{2 ∣ x - 3 ∣}{∣ 6 - x ∣} < ϵ

∣ x - 3 ∣< \frac{∣ 6 - x ∣ ϵ}{2}

then

∣ x - 3 ∣< 1 \Rightarrow 2 < x < 4 \Rightarrow - 4 < x < - 2 \Rightarrow 2 < 6 - x < 4 \Rightarrow 2 <∣ 6 - x ∣< 4

∣ x - 3 ∣< \frac{∣ 6 - x ∣ ϵ}{2} < \frac{2 ϵ}{2} = ϵ

then lets

δ = \min (1, ϵ)

if δ = 1, then 1 < ϵ \Rightarrow ϵ > 1

∣ x - 3 ∣< 1 \Rightarrow \frac{2 ∣ x - 3 ∣}{∣ 6 - x ∣} < \frac{2}{∣ 6 - x ∣}

2 < x < 4 \Rightarrow 2 < 6 - x < 4 \Rightarrow \frac{1}{4} < \frac{1}{6 - x} < \frac{1}{2} \Rightarrow \frac{1}{4} < \frac{1}{∣ 6 - x ∣} < \frac{1}{2}

\frac{2 ∣ x - 3 ∣}{∣ 6 - x ∣} < \frac{2}{∣ 6 - x ∣} < \frac{2}{2} < 1 < ϵ

∣ \frac{x}{6 - x} - 1 ∣< ϵ

then suppose that δ = ϵ

∣ x - 3 ∣< δ \Rightarrow \frac{2 ∣ x - 3 ∣}{∣ 6 - x ∣} < \frac{2 ϵ}{2} = ϵ

∣ \frac{x}{6 - x} - 1 ∣< ϵ

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