One-Sided Limit of Real Function/Examples/e^-1 over x at 0 from Left
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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:
\(\ds \lim_{x \mathop \to 0^-} \map f x\) | \(=\) | \(\ds +\infty\) |
Proof
By definition of the limit from the left:
- $\ds \lim_{x \mathop \to a^-} \map f x = A$
- $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: a - \delta < x < a \implies \size {\map f x - L} < \epsilon$
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:
\(\ds e^{1 / z}\) | \(>\) | \(\ds \epsilon\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds 1 / z\) | \(>\) | \(\ds \ln \epsilon\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds z\) | \(<\) | \(\ds \dfrac 1 {\ln \epsilon}\) |
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:
- $\ds \lim_{x \mathop \to 0^-} e^{-1 / x} = +\infty$
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $1 \, \text {(c)}$