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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!

Remy
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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 $f(x)=\underset{x \to x_0}\lim f(x_0)$).

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Chill2Macht
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    If we took f(x) = {cos(1/x) if x $\neq$ 0 and 0 if x = 0} would we prove that the limit DNE using the same argument? The image sequence from the left and the right converge to -1 but the constant sequence x = 0 converges to 0. Thus the limit DNE? – Remy Oct 26 '16 at 20:05
  • @JohnH Why do the image sequences from the left and right converge? What is the limit of $\cos(x)$ as $x \to \infty$? Consider the following sequences $x_n \to \infty$ (thus $\frac{1}{x_n} \to 0$): $x_0 \in [0,2 \pi)$, $x_n = x_{n-1} + 2\pi$. Then you can get constant sequences $\cos(x_n)$ converging to any value in $[0,1]$ as $x_n \to \infty$. Actually I am probably saying the exact same thing that you are -- anyway this should prove that the limit DNE yes. Related: http://math.stackexchange.com/a/834254/327486 – Chill2Macht Oct 26 '16 at 20:12
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    Sorry, I forgot to mention the domain isn't [0,1] in this case, We are looking at all x. And doesn't cos(1/x) converge to -1 from the left and right though? I am confused. – Remy Oct 26 '16 at 20:18
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    Ok, I will take a look at that thread. – Remy Oct 26 '16 at 20:19
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    I am having difficulty understanding the solutions in that thread. I want to see if I can't prove this using the same argument as above. I am posting a new question right now. – Remy Oct 26 '16 at 20:43