How do you prove that the limit of this function does not exist? So I've been trying to figure out how this professor did this proof in which he had to prove that the limit of $sin\frac{π}{2x}$ as $x$ approaches 0 does not exist.
Could someone go over the steps he did and why he was able to do what he did?
https://imgur.com/RAUrwRK
 A: First, they show if we take $\varepsilon = 1$, than given any $\delta > 0$ we can find $y,z \in (0,\delta)$ such that $h(z) = 1, h(y) = -1$.
This fact shows there can't be $L \in \mathbb{R}$, such that $L$ is the limit (because give $\varepsilon = 1$, there exists no $\delta > 0$ such that $|x|<\delta \Rightarrow |h(x)-L|< \varepsilon = 1$).
A: Let
$f(x)
=\sin\frac{π}{2x}
$.
Let $n$ be a positive integer.
If
$x = \frac1{2n}$,
then
$f(x)
=\sin\frac{π}{2(1/(2n))}
=\sin(\pi n)
=0
$.
If
$x = \frac1{4n+1}$,
then
$f(x)
=\sin\frac{π}{2(1/(4n+1))}
=\sin(\pi (2n+\frac12))
=1
$.
For these two sequences 
approaching $0$,
the limits exists and
are different.
Therefore tha function
has no limit as
$x \to 0$.
A: To show that the limit does not exist it is enough to show that $\lim {x \rightarrow 0^+}$ does not exist.
$z:= \dfrac {π}{2x}$.
Consider $\lim_{z \rightarrow  + \infty } \sin(z)$.
1) Choose $z_n = nπ,$  $n \in \mathbb{N}.$
$y_n:= \sin (z_n) = 0,$ for all $n$.
2) Choose $z_n = π/2 +2πn$.
$y_n= \sin(z_n) = 1$, for all $n$.
1)$ \lim_{n \rightarrow \infty} y_n = 0.$
2) $\lim_{n \rightarrow \infty} y_n = 1.$
Limit does not exist.
