0
$\begingroup$

enter image description here

I know because rational and irrational numbers are dense in R, then for any $x$, you can always find both a sequence of Rationals and a sequence of Irrationals that converge to $x$

Qustions:

1)Why do we choose sequence of rational numbers for irrational x(sequence converges to irrational x)

2)Why in the first case $1 = \lim f(x_n)$ and in the seond $0 = \lim f(y_n)$

2 Answers 2

1

1) We choose a sequence of rational numbers because the function is $1$ at all rational numbers, thus $\lim_{n\to\infty} f(x_n)=1$ (more on that below).

It would also be possible to prove discontinuity using the $\epsilon-\delta$ definition, but this particular textbook chose the convergent sequences way of doing it.

2) In the first case, for every value of $n$, the value $f(x_n)$ is equal to $1$, which means that $$\lim_{n\to\infty} f(x_n)=\lim_{n\to\infty} 1 = 1.$$ A similar case can be made for $0=\lim f(y_n)$, as $f(y_n)=0$ for all values of $n$.

  • 0
    Great, everything is clear right now! Thank you! P.S if you have posibility can you please show the proof using $\epsilon - \delta$ definition? – 2017-02-10
  • 2
    @DanielYefimov I suggest you first try to do the proof yourself. After all, that's the only way we learn anything in math: my trying it on our own. (if you fail, you can still post a question and someone, maybe even me, will answer it. But try it yourself, first!) – 2017-02-10
0

1) because it helps to prove the desired discontinuity

2) because both sequences are constant ones !