Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 62747 by Mikael last updated on 24/Jun/19

Are f, g: R→R defined by  f(x)= { ((0,   x ∈ R\Q)),((x,   x ∈Q)) :}  g(x)= { ((1,   x=0)),((0,    x≠0)) :}  show that lim_(x→0)  f(x)=0 and lim_(y→0)  g(y)=0  however lim_(x→0)  g(f(x)) does not exist.

$${Are}\:\boldsymbol{{f}},\:\boldsymbol{{g}}:\:\mathbb{R}\rightarrow\mathbb{R}\:{defined}\:{by} \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{0},\:\:\:{x}\:\in\:\mathbb{R}\backslash\mathbb{Q}}\\{{x},\:\:\:{x}\:\in\mathbb{Q}}\end{cases} \\ $$$${g}\left({x}\right)=\begin{cases}{\mathrm{1},\:\:\:{x}=\mathrm{0}}\\{\mathrm{0},\:\:\:\:{x}\neq\mathrm{0}}\end{cases} \\ $$$${show}\:{that}\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\mathrm{0}\:{and}\:\underset{\boldsymbol{\mathrm{y}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{y}}\right)=\mathrm{0} \\ $$$${however}\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\boldsymbol{\mathrm{lim}}}\:\boldsymbol{\mathrm{g}}\left(\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)\right)\:{does}\:{not}\:{exist}. \\ $$

Answered by MJS last updated on 25/Jun/19

x∈R\Q: lim_(x→0)  f(x)=0 is trivial because f(x)=0 in this case  x∈Q: lim_(x→0)  f(x)=lim_(p→0)  (p/q) ∀p∈Z∧∀q∈N^★  ⇒  ⇒ lim_(x→0)  f(x)=lim_(p→0)  (p/q)=0  lim_(x→0)  g(x)=0 because g(x)=0 for x∈[−ε; ε]∀ε∈R^+   lim_(x→0)  g(f(x)) does not exist because ∀x∈R\Q  g(f(x))=1 but ∀x∈Q g(f(x))=0 ⇒   ⇒ lim_(x∈R\Q→0) g(f(x))=1 ∧ lim_(x∈Q→0) g(f(x))=0 ⇒  limit ∀x∈R doesn′t exist

$${x}\in\mathbb{R}\backslash\mathbb{Q}:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\mathrm{0}\:\mathrm{is}\:\mathrm{trivial}\:\mathrm{because}\:{f}\left({x}\right)=\mathrm{0}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$${x}\in\mathbb{Q}:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{p}}{{q}}\:\forall{p}\in\mathbb{Z}\wedge\forall{q}\in\mathbb{N}^{\bigstar} \:\Rightarrow \\ $$$$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}\right)=\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{p}}{{q}}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{g}\left({x}\right)=\mathrm{0}\:\mathrm{because}\:{g}\left({x}\right)=\mathrm{0}\:\mathrm{for}\:{x}\in\left[−\epsilon;\:\epsilon\right]\forall\epsilon\in\mathbb{R}^{+} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{g}\left({f}\left({x}\right)\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{exist}\:\mathrm{because}\:\forall{x}\in\mathbb{R}\backslash\mathbb{Q} \\ $$$${g}\left({f}\left({x}\right)\right)=\mathrm{1}\:\mathrm{but}\:\forall{x}\in\mathbb{Q}\:{g}\left({f}\left({x}\right)\right)=\mathrm{0}\:\Rightarrow\: \\ $$$$\Rightarrow\:\underset{{x}\in\mathbb{R}\backslash\mathbb{Q}\rightarrow\mathrm{0}} {\mathrm{lim}}{g}\left({f}\left({x}\right)\right)=\mathrm{1}\:\wedge\:\underset{{x}\in\mathbb{Q}\rightarrow\mathrm{0}} {\mathrm{lim}}{g}\left({f}\left({x}\right)\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{limit}\:\forall{x}\in\mathbb{R}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$

Commented by Mikael last updated on 25/Jun/19

thank you Sir.

$${thank}\:{you}\:{Sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com