Both $x_n$ and $y_n$ are sequences in (0, 1) and they both converge to 0.
Both image sequences {f($x_n$)} and {f($y_n$)} converge, but they converge to different numbers.
Prove that it must be the case that $\lim_{x\to 0}$ f(x) does not exist (i.e. for any number L, it is not the case that $\lim_{x\to 0}$f(x) = L)
This is a definition I could try to make use of: $\lim_{x\to a}$f (x) = L means that:
$(∀ε > 0)(∃δ > 0)(∀x ∈ D_f )[0 < |x − a| < δ \Rightarrow |f (x) − L| < ε].$
An additional Theorem I could make use of is $\lim_{x\to a}$f (x) = L if and only if:
For every sequence $x_k$ in $D_f$ \ {a}, if $x_k$ → a, then f($x_k$) → L.
My current solution: Since $x_n$ and $y_n$ both converge to 0, we can denote 0 as 'a' so $x_n,y_n$ → a. However, f($x_n$) and f($y_n$) both converge to different numbers so it cannot be the case that they both converge to L. This completes the proof.
Any corrections or suggestions to make my proof more elegant would be appreciated!
Best Answer
Your argument is correct.
If there existed a number $L$ such that $\underset{x \to 0}\lim f(x) =L$, then $f(x_n) \to L$ for all sequences $x_n \to 0$, but $L$ can't be equal to two distinct numbers, hence the contradiction.
As an aside, this problem is related to the issue of sequential continuity and the various possible characterizations of continuity for real functions.
By assumption, because $x_n$ and $y_n$ converge to the same value, but $f(x_n)$ and $f(y_n)$ don't, $f$ isn't sequentially continuous at $0$. (See here.) At least for $\mathbb{R}$, regular continuity implies sequential continuity -- see for example here -- thus the absence of sequential continuity implies the absence of regular continuity by contraposition. Since $f$ isn't continuous at $0$, we must have either that $f(x) \not= \underset{x \to 0}\lim f(x)$ or that $\underset{x \to 0}\lim f(x)$ does not exist. (Since a function $f$ is continuous at a point $x_0$ if and only if $f(x)=\underset{x \to x_0}\lim f(x_0)$).