One-Sided Limit of Real Function/Examples/e^-1 over x at 0 from Right

Example of One-Sided Limit of Real Functions
Let $f: \R \to \R$ be the real function defined as:
 * $\map f x = e^{-1 / x}$

Then:

Proof

 * Limit-of-e-to-minus-1-over-x.png

By definition of the limit from the right:
 * $\ds \lim_{x \mathop \to a^+} \map f x = A$


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: a < x < a + \delta \implies \size {\map f x - L} < \epsilon$
 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: a < x < a + \delta \implies \size {\map f x - L} < \epsilon$

In this case we are interested in the situation where $a = 0$, and we wish to demonstrate that $L = 0$ at that point.

Hence the condition we need to ascertain is:


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: 0 < x < \delta \implies \size {e^{-1 / x} } < \epsilon$

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

Let $x > 0$.

We have that:
 * $e^{-1 / x} > 0$

and so:
 * $\size {e^{-1 / x} } = e^{-1 / x}$

Then we have:

So, having been given an arbitrary $\epsilon \in \R_{>0}$, let $\delta = \dfrac 1 {\map \ln {1 / \epsilon} }$.

Then:
 * $0 < x < \delta \implies \size {e^{-1 / x} } < \epsilon$

Hence by definition of limit from the right:
 * $\ds \lim_{x \mathop \to 0^+} e^{-1 / x} = 0$