Limit with Epsilon Powers of 2

Theorem
Let $\openint a b$ be an open real interval.

Let $c \in \openint a b$.

Let $f: \openint a b \setminus \set c \to \R$ be a real function.

Let $L \in \R$.

Suppose that:


 * $\forall n > 0 \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $

Then the limit of $f$ exists as $x$ tends to $c$, and is equal to $L$.

Proof
Denote by $\map P {2^{-n},\delta}$ the proposition considered in the theorem exposition:


 * $\forall \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $

Let the limit of $f$ as $x \to c$ exist and equal $L$, as described in the definition of limit:


 * $\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$.

Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds.

From the Archimedean Principle:


 * $\forall \epsilon > 0: \exists n \in \N: n > \dfrac 1 \epsilon$.

As $2^n > n$ for all $n$, we have that $2^n > \dfrac 1 \epsilon$.

By Ordering of Reciprocals, $2^{-n} < \epsilon$.

Therefore $\size {\map f x - L} < \epsilon$ is implied by $\size {\map f x - L} < 2^{-n}$.

Thus $\map P {\epsilon,\delta}$ is a weaker statement $\map P {2^{-n},\delta}$