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.

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

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

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 $\displaystyle x\rightarrow{1}$

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

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

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

Hence f is discontinuous at x=1

When $\displaystyle x\rightarrow{2}$

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

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

$\displaystyle Also~f(2)=20$

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

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.

Your generosity is greatly appreciated.