Limit of Real Function/Examples/Identity Function with Discontinuity at 0

Example of Limit of Real Function
Let $f$ be the real function defined as:
 * $\map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$

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

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


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

In this instance, we have:


 * $\map f x = x$ for $x \ne 0$
 * $A = 0$

Let $\epsilon \in \R_{>0}$ be arbitrary.

Let $\delta = \epsilon$.

Let $0 < \size {x - 1} < \delta = \epsilon$.

Then we obtain:

Hence the result.