Limit of Real Function/Examples/Sine of Reciprocal of x at 0

Example of Limit of Real Function
Let:
 * $\map f x = \map \sin {\dfrac 1 x}$

Then:
 * $\ds \lim_{x \mathop \to 0} \map f x$

does not exist.

Proof
By definition of the limit of a real function:
 * $\ds \lim_{x \mathop \to a} \map f x = A$


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$

Let $\epsilon = \dfrac 1 2$.

Let $\delta \in \R_{>0}$.

Let $n \in \N$ such that $N = \dfrac 2 {\pi \paren {1 + 4 n} } < \delta$.

Then:
 * $\map \sin N = 1$

and so it is not the case that:
 * $0 \size x < \delta \implies \size {\map f x} < \epsilon$

It follows that $\map \sin {\dfrac 1 x}$ has no limit at $x - 0$.