# Continuity Problem.

• May 31st 2010, 05:07 PM
Sudharaka
Continuity Problem.
Hi everyone,

I need your idea about the way in which I had done the following question. Although this could be proved using the "sequential criterion of continuity" I want to know whether my way of doing it is correct or not.

$Let~f:R\rightarrow{R},~where$

$f(x)=\left\{\begin{array}{cc}5x^2 &\mbox{ if }x\in{Q}\\x^3+6x&\mbox{ if }x\in{P}\end{array}\right.$

Note:Q denotes rational numbers and P denotes irrational numbers.

Show that f is discontinuous at x=1 and continuous at x=2. Is it true that f is continuous only at x=2? Justify your answer.

Solution:

When $x\rightarrow{1}$

$If~x\in{Q}\Rightarrow{\lim_{x\rightarrow{1}}f(x)}= \lim_{x\rightarrow{1}}5x^2=5$

$If~x\in{P}\Rightarrow{\lim_{x\rightarrow{1}}f(x)}= \lim_{x\rightarrow{1}}x^3+6x=7$

$Therefore~\lim_{x\rightarrow{1}}f(x)~does~not~exsi t.$

Hence f is discontinuous at x=1

When $x\rightarrow{2}$

$If~x\in{Q}\Rightarrow{\lim_{x\rightarrow{2}}f(x)}= \lim_{x\rightarrow{2}}5x^2=20$

$If~x\in{P}\Rightarrow{\lim_{x\rightarrow{2}}f(x)}= \lim_{x\rightarrow{2}}x^3+6x=20$

$Also~f(2)=20$

$Therefore~\lim_{x\rightarrow{2}}f(x)=f(2)$

Hence by the 3-point definition of continuity f(x) is continuous at x=2.

Is it true that f is continuous only at x=2?

No. Consider x=0; It could be shown using the previous method that x is continuous at x=0.

• May 31st 2010, 06:05 PM
tonio
Quote:

Originally Posted by Sudharaka
Hi everyone,

I need your idea about the way in which I had done the following question. Although this could be proved using the "sequential criterion of continuity" I want to know whether my way of doing it is correct or not.

$Let~f:R\rightarrow{R},~where$

$f(x)=\left\{\begin{array}{cc}5x^2 &\mbox{ if }x\in{Q}\\x^3+6x&\mbox{ if }x\in{P}\end{array}\right.$

Note:Q denotes rational numbers and P denotes irrational numbers.

Show that f is discontinuous at x=1 and continuous at x=2. Is it true that f is continuous only at x=2? Justify your answer.

Solution:

When $x\rightarrow{1}$

$If~x\in{Q}\Rightarrow{\lim_{x\rightarrow{1}}f(x)}= \lim_{x\rightarrow{1}}5x^2=5$

$If~x\in{P}\Rightarrow{\lim_{x\rightarrow{1}}f(x)}= \lim_{x\rightarrow{1}}x^3+6x=7$

$Therefore~\lim_{x\rightarrow{1}}f(x)~does~not~exsi t.$

Hence f is discontinuous at x=1

When $x\rightarrow{2}$

$If~x\in{Q}\Rightarrow{\lim_{x\rightarrow{2}}f(x)}= \lim_{x\rightarrow{2}}5x^2=20$

$If~x\in{P}\Rightarrow{\lim_{x\rightarrow{2}}f(x)}= \lim_{x\rightarrow{2}}x^3+6x=20$

$Also~f(2)=20$

$Therefore~\lim_{x\rightarrow{2}}f(x)=f(2)$

Hence by the 3-point definition of continuity f(x) is continuous at x=2.

3-point def. of cont.? What's that?

Is it true that f is continuous only at x=2?

No. Consider x=0; It could be shown using the previous method that x is continuous at x=0.

It is continuous also at $x=3$ ...(Wink)

Tonio

.
• May 31st 2010, 08:46 PM
Drexel28
Quote:

Originally Posted by Sudharaka
Hi everyone,

I need your idea about the way in which I had done the following question. Although this could be proved using the "sequential criterion of continuity" I want to know whether my way of doing it is correct or not.

$Let~f:R\rightarrow{R},~where$

$f(x)=\left\{\begin{array}{cc}5x^2 &\mbox{ if }x\in{Q}\\x^3+6x&\mbox{ if }x\in{P}\end{array}\right.$

Note:Q denotes rational numbers and P denotes irrational numbers.

Show that f is discontinuous at x=1 and continuous at x=2. Is it true that f is continuous only at x=2? Justify your answer.

There's a really nice method for these piecewise functions defined on the rationals and irrationals. Let $\{q_n\},\{i_n\}$ be rational, irrational sequences respectively such that $\lim\text{ }q_n=\lim\text{ }i_n=x_0$ then by continuity $f(x_0)=\lim\text{ }f(q_n)=\lim\text{ }5q_n^2=5x_0^2$ but with equal validity $f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }\left(i_n^3+6i_n\right)=x_0^3+6x_0$ from where it follows that $f\text{ is continuous at }x_0\implies 5x_0^2=x_0^3+6x_0$ and so you must only check continuity at the solutions to that equation.
• May 31st 2010, 11:05 PM
Sudharaka
Quote:

Originally Posted by Drexel28
There's a really nice method for these piecewise functions defined on the rationals and irrationals. Let $\{q_n\},\{i_n\}$ be rational, irrational sequences respectively such that $\lim\text{ }q_n=\lim\text{ }i_n=x_0$ then by continuity $f(x_0)=\lim\text{ }f(q_n)=\lim\text{ }5q_n^2=5x_0^2$ but with equal validity $f(x_0)=\lim\text{ }f(i_n)=\lim\text{ }\left(i_n^3+6i_n\right)=x_0^3+6x_0$ from where it follows that $f\text{ is continuous at }x_0\implies 5x_0^2=x_0^3+6x_0$ and so you must only check continuity at the solutions to that equation.

Dear Drexel28,

Thank you very much for the reply. But can you tell me weather my method is correct or not? And there's another problem which I came across when reading your answer. How do we know that such sequences $\{q_n\},\{i_n\}$ exist in the domain of f ?

What I meant by the 3-point defintion is, the "limit definition of continuity".
$\left(i.e~ f~ is~ continuous~ at~ x=c\leftrightarrow{\lim_{x\rightarrow{c}}f(x)=f(c) }\right)$

Sorry if it caused you confusion.
• Jun 1st 2010, 01:46 PM
Drexel28
Quote:

Originally Posted by Sudharaka
Dear Drexel28,

Thank you very much for the reply. But can you tell me weather my method is correct or not? And there's another problem which I came across when reading your answer. How do we know that such sequences $\{q_n\},\{i_n\}$ exist in the domain of f ?

What I meant by the 3-point defintion is, the "limit definition of continuity".
$\left(i.e~ f~ is~ continuous~ at~ x=c\leftrightarrow{\lim_{x\rightarrow{c}}f(x)=f(c) }\right)$

Sorry if it caused you confusion.

Your method worked except you failed to check the last root of $x^3-5x^2+6x=0$. We know such sequences existence by the density of the rationals and irrationals in the reals.