Prove continuity

[cainemvhzc]

7. BONUS (4 points) Let f be the function defined by:

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}x^2&\mbox{ if }\, x\, \mbox{ is rational}\\-x^2&\mbox{ if }\, x\, \mbox{ is irrational}\end{cases}\)

Is f continuous at x = 0? If so, prove it. If not, prove that it is not.

 

[pka]

7. BONUS (4 points) Let f be the function defined by:

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}x^2&\mbox{ if }\, x\, \mbox{ is rational}\\-x^2&\mbox{ if }\, x\, \mbox{ is irrational}\end{cases}\)

Is f continuous at x = 0? If so, prove it. If not, prove that it is not.

You must show us some effort you have done. Then say where you have difficulty.

Hints: Is \(\displaystyle f\) defined at \(\displaystyle x\,=\,0~? \)

\(\displaystyle \varepsilon \,> \,0\; \Rightarrow \;\delta = \sqrt \varepsilon \).

 

[cainemvhzc]

You must show us some effort you have done. Then say where you have difficulty.

Hints: Is \(\displaystyle f\) defined at \(\displaystyle x\,=\,0~? \)

\(\displaystyle \varepsilon \,> \,0\; \Rightarrow \;\delta = \sqrt \varepsilon \).

Can it only be done by delta-epsilon method?
How about plugging 0 in 0^- and 0⁺ ? Will that be wrong?

 

[daon2]

Can it only be done by delta-epsilon method?
How about plugging 0 in 0^- and 0⁺ ? Will that be wrong?

Edit: Whoops, I misread.

The big question is if the limit from the left and right both exist, and if they do, are they equal?

 

[pka]

Can it only be done by delta-epsilon method?
How about plugging 0 in 0^- and 0⁺ ? Will that be wrong?

Do you understand \(\displaystyle |-x^2|=|0-x^2|=|x^2-0|=|x^2|=x^2~?\)

Now if \(\displaystyle |x|<\delta\) then \(\displaystyle |x^2-0|=|x^2|<\varepsilon \)

In this case two sides do not matter.