piecewise limit (help on rational/irrational)

[letsgetaway]

I don't know how to figure out this type of piecewise limit. I understand what a rational and irrational number is, but I don't know how to apply it to this problem. There is a closed bracket typo at the end of the function.

\(\displaystyle \begin{array}{l}
\lim f(x) = \left\{ \begin{array}{l}
2x, rational \\
x, irrational \\
\end{array} \\
x -> 1 \\
\end{array}
\]\)

 

[pka]

This function does not have a limit at any non-zero number.
To see this, recall that between any two numbers there are both rational and irrational numbers. Therefore, if \(\displaystyle \delta > 0,\;\left( {\exists \eta \in Q} \right)\left( {\exists \lambda \in R\backslash Q} \right)\left[ {\{ \lambda ,\eta \} \subseteq (1 ,1 + \delta )} \right]\) so that \(\displaystyle \left| {f(\lambda ) - f(\eta )} \right| = \left| {2\lambda - \eta } \right| \ge 1.\)
Thus, the limit cannot exist.

 

[soroban]

Hello, letsgetaway!

I don't know how to figure out this type of piecewise limit.
I understand what a rational and irrational number is,
but I don't know how to apply it to this problem.

\(\displaystyle \L\;\;\lim_{x\to1}f(x)\:=\:\begin{Bmatrix}2x, & x\text{ rational} \\ x, & x\text{ irrational} \end{Bmatrix}\)

We know that: \(\displaystyle \,f(1)\;=\;2\)

But the function is discontinuous at every point!

If such a graph could be drawn, it might look like this:

          |
          |           *
          |
          |         *
          |
          |       *
          |               *
        2 +     o       *
          |           *
          |   *     *
          |       *
          | *  *
          |  *
      - - * - - + - - - - -
          |     1

I agree with pka . . . we cannot find the limit as \(\displaystyle x\to1\)

Between any two irrational numbers, there is a rational number.
Between any two rational numbers, there is an irrational number.

So that rational and irrational numbers are alternately "interwoven" on the number line,
\(\displaystyle \;\;\)despite the fact that there are more irrational numbers than rational numbers
\(\displaystyle \;\;\)(but that's yet another story . . . )

 

[pka]

The function is actually continuous at zero.

If \(\displaystyle \varepsilon > 0\) then define \(\displaystyle \delta = \frac{\varepsilon }{2}.\)
So \(\displaystyle \left| {x - 0} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 0} \right| < \varepsilon .\)
Because \(\displaystyle \begin{array}{rcl}
x \in Q\quad & \Rightarrow & \quad \left| {f(x) - 0} \right| = \left| x \right| < \delta < \varepsilon \\
x \in R\backslash Q \quad & \Rightarrow & \quad \left| {f(x) - 0} \right| = \left| {2x} \right| < 2\delta = \varepsilon \\
\end{array}\)