show-that-lim-x-0-x-does-not-exist-Hence-define-x-and-sketch-a-graph-for-y-3x-2-x-

Question Number 72806 by Rio Michael last updated on 03/Nov/19

\underset{x \to 0}{s h o w t h a t \lim} [x] d o e s n o t e x i s t .

H e n c e d e f i n e [x] a n d s k e t c h a g r a p h f o r

y = 3 x^{2} + [x]

Commented by mathmax by abdo last updated on 03/Nov/19

l e t f (x) = [x] w e h a v e {l i m}_{x \to 0^{+}} f (x) = {l i m}_{x \to 0^{+}} [x] = 0

{l i m}_{x \to 0^{-}} f (x) = {l i m}_{x \to 0^{-}} [x] = - 1 a n d f (0) = 0 w e h a v e 0 \neq - 1 \Rightarrow

{l i m}_{x \to 0} [x] d o n t e x i s t

Commented by Rio Michael last updated on 03/Nov/19

l o v e t h a t m e t h o d, t h a n k s s i r

Answered by mind is power last updated on 03/Nov/19

suppose that \lim_{x \to 0} [x] exist

call l = \lim_{x \to 0} [x]

\Rightarrow \forall ε > 0 \exists η > 0 ∣ ∣ x ∣< η \Rightarrow∣ [x] - l ∣< ε

let ϵ = \frac{l}{3}

\Rightarrow \exists η_{ε} such that \forall x \in] - η, η [∣ [x] - L ∣< \frac{1}{3}

for x \in] - η, 0 [\Rightarrow∣ - 1 - L ∣< \frac{1}{2} \Leftrightarrow∣ 1 + L ∣< \frac{1}{3}

for x \in] 0, η [\Rightarrow∣ - L ∣< \frac{1}{3}

we have ∣ 1 + L - L ∣= 1

truangular inquality \Rightarrow∣ 1 ∣= 1 ⩽∣ - L ∣ + ∣ 1 + L ∣⩽ \frac{1}{3} + \frac{1}{3} = \frac{2}{3}

absurd since that L / dosent exist

2)

y = {3 x}^{2} + k x \in [k, k + 1 [

Commented by Rio Michael last updated on 03/Nov/19

t h a n k s

Commented by mind is power last updated on 03/Nov/19

most Welcom