Existence of Functional Limits

This is a corollary from the Sequential Criterion For Functional Limits:

Let $f : A \to R$ where $A \subseteq R$ and $c$ is a Limit Point of $A$ .
If $\exists$ sequences $(x_{n}) \subseteq A - {c}$ and $(y_{n}) \subseteq A - {c}$ with $x_{n} \to c, y_{n} \to c$ but $(f (x_{n}))$ and $(f (y_{n}))$ do not converge to the same value. Then:

lim_{x \to c} f (x)

does not exist.

Namely this is the opposite of the Sequential Criterion For Functional Limits (2).

Example of Finding Functional Limits

Let $f (x) = \sin (\frac{1}{x})$ where $A = R - {0}$ .

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f(x) = sin(1/x)

A reasonable question to ask is if/what $lim_{x \to 0} f (x)$ is here (notice $0$ here is a limit point of the domain set). But notice that the sequence of points that intersect the graph at 0 on the right hand side make up a sequence where $f (x_{n}) \to 0$ . But we can do the same thing with the points that intersect the line $y = 1$ . Then $f (y_{n}) \to 1$ , so we have two sequences that converge to different limits, so then the limit of the original function must not exist.

More explicitly, let $n \in N$ . Then choose:

x_{n} = \frac{1}{n π}, y_{n} = \frac{1}{\frac{π}{2} + 2 n π}

Then notice that $x_{n}, y_{n} \to 0$ individually, but:

f (x_{n}) = \sin (\frac{1}{\frac{1}{n π}}) = \sin (n π) = 0

f (y_{n}) = \sin (\frac{1}{\frac{1}{\frac{π}{2} + 2 n π}}) = \sin (\frac{π}{2} + 2 n π) = 1

So since $f (x_{n}) \to 0$ and $f (y_{n}) \to 1$ then by our corrolary above then we get the limit does not exist.