Power Function on Base greater than One tends to One as Power tends to Zero/Rational Number/Lemma

Lemma
Let $a \in \R$.

Let $x \in \R$.

Let $a > 1$.

Let $0 < x < 1$.

Then $1 < a^x < 1 + ax$.

Proof
Define a real mapping $g_x: \R_{> 0} \to \R$ as:
 * $g_x \left({ a }\right) = 1 + ax - a^x$

Then:
 * $D_a g_x = x \left({ 1 - a^{x-1} }\right)$

We show now that the derivative of $g$ is positive for all $a > 1$:

So $D_a g_x$ is increasing for all $a > 1$ and $0 < x < 1$.

Further, $D_a g_x \left({ 1 }\right) = 0$.

Thus $D_a g \left({ a }\right)$ is positive for all $a > 1$.

Whence, by Derivative of Monotone Function, $g$ is increasing for all $a > 1$.

Now, $g \left({ 1 }\right) = x > 0$.

So $g \left({ a }\right)$ is positive for all $a > 1$.

That is:
 * $1 + ax - a^x > 0$

From Real Number Ordering is Compatible with Addition:
 * $a^x < 1 + ax$

Now:

So, for $0 < x < 1$:
 * $1 < a^x < 1 + ax$