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

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Theorem

Let $a \in \R$ be a real number such that $a > 1$.

Let $r \in \Q_{> 0}$ be a strictly positive rational number such that $r < 1$.


Then:

$1 < a^r < 1 + a r$



Proof

Define a real function $g_r: \R_{> 0} \to \R$ as:

$\map {g_r} a = 1 + a r - a^r$

Then differentiating with respect to $a$ gives:

$D_a \map {g_r} a = r \paren {1 - a^{r - 1} }$


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

\(\displaystyle r\) \(<\) \(\displaystyle 1\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle r - 1\) \(<\) \(\displaystyle 0\) Subtract $1$ from both sides
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{r - 1}\) \(<\) \(\displaystyle a^0\) Power Function on Base Greater than One is Strictly Increasing: Rational Number
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^{r - 1}\) \(<\) \(\displaystyle 1\) Definition of Integer Power
\(\displaystyle \leadsto \ \ \) \(\displaystyle -1\) \(<\) \(\displaystyle -a^{r - 1}\) Order of Real Numbers is Dual of Order of their Negatives
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle 1 - a^{r - 1}\) adding $1$ to both sides
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle r \left({1 - a^{r - 1} }\right)\) multiplying both sides by $r > 0$
\(\displaystyle \leadsto \ \ \) \(\displaystyle 0\) \(<\) \(\displaystyle D_a \map {g_r} a\) from the formula found for $D_a \map {g_r} a$ above


So $D_a \map {g_r} a$ is positive for all $a > 1$.

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


Now:

$\map {g_r} 1 = r > 0$

So $\map {g_r} a$ is positive for all $a > 1$.

That is:

$0 < 1 + a r - a^r$

Adding $a^r$ to both sides of the above yields:

$a^r < 1 + a r$


Finally:

\(\displaystyle 0\) \(<\) \(\displaystyle r\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^0\) \(<\) \(\displaystyle a^r\) Power Function on Base Greater than One is Strictly Increasing: Rational Number
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(<\) \(\displaystyle a^r\) Definition of Integer Power


So, for $0 < r < 1$:

$1 < a^r < 1 + a r$

$\blacksquare$