Limit of Real Function/Examples/e^-1 over size x at 0

Example of Limit of Real Function

$\ds \lim_{x \mathop \to 0} e^{-1 / \size x} = 0$

Proof

By definition of the limit of a real function:

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

if and only if:

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

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

Then we have:

\(\ds e^{-1 / \size x}\)

\(<\)

\(\ds \epsilon\)

\(\ds \leadstoandfrom \ \ \)

\(\ds -1 / \size x\)

\(<\)

\(\ds \ln \epsilon\)

\(\ds \leadstoandfrom \ \ \)

\(\ds 1 / \size x\)

\(>\)

\(\ds -\ln \epsilon\)

\(\ds \)

\(=\)

\(\ds \map \ln {\dfrac 1 \epsilon}\)

Logarithm of Reciprocal

\(\ds \leadstoandfrom \ \ \)

\(\ds \size x\)

\(<\)

\(\ds \dfrac 1 {\map \ln {1 / \epsilon} }\)

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

Then:

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

Hence by definition of limit of a real function:

$\ds \lim_{x \mathop \to 0} e^{-1 / \size x} = 0$

$\blacksquare$

Sources

• 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $1 \, \text {(a)}$