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

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 left:
 * $\ds \lim_{x \mathop \to a^-} \map f x = A$


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

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

Hence the condition we need to ascertain is:


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

That is:


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

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

Let $\epsilon \in \R_{>0}$ be chosen arbitrarily such that $\epsilon < 1$.

Let $x < 0$.

Let $z = -x$.

Thus:
 * $z > 0$

Then we have:

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

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

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