ProofWiki:Sandbox

Theorem
Let $a \in \R_{> 0}$.

Let $a > 1$.

Let $f : \Q \to \R$ be the real mapping defined as:
 * $f \left({ q }\right) = a^q$

where $a^q$ denotes $a$ to the power of $q$.

Then:
 * $\displaystyle \lim_{x \to 0} f \left({ x }\right) = 1$

Proof
We start by treating the right-sided limit.

Let $0 < x < 1$

From the lemma:
 * $1 < a^x < 1 + ax$

Also:

So from the Squeeze Theorem:
 * $\displaystyle \lim_{ x \to 0^{+} } a^x = 1$

We now treat the left-sided limit.

Let $-1 < x < 0$.

Also:

So from the Squeeze Theorem:
 * $\displaystyle \lim_{ x \to 0^{-} } a^x = 1$

From Limit iff Limits from Left and Right:
 * $\displaystyle \lim_{ x \to 0 } a^x = 1$

Hence the result.