Find a limit of a weird function

Question

We have this function:

$ f(x) = \begin{cases} 0, & x \in \mathbb{Z} \\[2ex] 1, & x \in \mathbb{R} \backslash \mathbb{Z} \end{cases} $

Find if these limits exist:

$ \lim_{x\to \infty} f(x) $

$ \lim_{x\to 2} f(x)$

$ \lim_{x\to 2.5} f(x) $

If not, prove it.

It's clear that there's a limit when $ x \to 2.5 $ but I'm having trouble proving it using the defnition of the limit,

any help?

thanks :)

Answer 1

Partial answer

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$\lim\limits_{x\to\infty}f(x)$ does not exist because:

• $\forall x : f(x)=0 \implies f(\lceil{x}\rceil)=1$
• $\forall x : f(x)=1 \implies f(x+\frac12)=0$

In other words, $\lim\limits_{x\to\infty}f(x)$ does not exist because:

• For every $a$ such that $f(a)=0$ there exists $b>a$ such that $f(b)=1$
• For every $a$ such that $f(a)=1$ there exists $b>a$ such that $f(b)=0$

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$\lim\limits_{x\to2.5}f(x)=0$ because:

• $\forall\epsilon:|\epsilon|<0.5 \implies f(2.5+\epsilon)=f(2.5-\epsilon)=0$

In other words, $\lim\limits_{x\to2.5}f(x)=0$ because:

• For every $\epsilon$ close enough to zero, $f(2.5+\epsilon)=f(2.5-\epsilon)=0$

Answer 2

2) The limit at $2$ is $0$ as all values around $2$ are $0$.

3) The limit at $2.5$ is $0$ as all values around $2.5$ are $0$.

1) There is no limit at $\infty$ as there are infinitely many $0$'s and $1$'s around $\infty$.