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

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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:

$\ds \lim_{x \mathop \to a} \map f x = A$

if and only if:

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


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:

$\sin N = 1$

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 $x = 0$.

$\blacksquare$


Sources

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