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:

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

if and only if:

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

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:

\(\ds \size {\map f x - 0}\)

\(=\)

\(\ds \size x\)

\(\ds \)

\(<\)

\(\ds \epsilon\)

Hence the result.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: $\S 1.3$: Limits of functions: Example $1.3.2$