Number to zero power calculus/ limit approach difficulty

Question

I'm trying to prove that $a^0 = 1$ using the calculus epsilon-delta limiting approach, i.e. I'm trying to prove the following

$$ \lim_{x\rightarrow 0} a^x = 1 $$

I'm not having much luck. At least part of the problem comes at the following point

$ |a^x - 1| < \varepsilon \Rightarrow \log_a(1-\varepsilon) < x < \log_a(1+\varepsilon) $

But then I can't write that as an inequality of the form

$-\delta(\varepsilon) < x < \delta(\varepsilon) \Rightarrow |x| < \delta(\varepsilon) $

Since

$-\log_a(1+\varepsilon) \neq \log_a(1-\varepsilon) $

(Or vice versa).

Help?

Note that, furthermore, if we say choose the function $\delta(\epsilon) = \log_a(1+\epsilon)$, we get

$-\log_a(1+\epsilon) < x < \log_a(1+\epsilon) \Rightarrow \frac{1}{1+\epsilon} < a^x < 1+\epsilon $

Which I can rewrite as

$ \frac{-\epsilon}{1+\epsilon} < a^x - 1 < \epsilon $

but I don't see ANY way of rewriting it as

$ |a^x - 1 | < \epsilon$

So...yeah

Answer 1

METHODOLOGY $1$: ELEMENTARY/PRE-CALCULUS APPROACH

Herein we present an approach that relies only on an elementary pair of inequalities and the squeeze theorem. To that end, we begin with the following primer.

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PRIMER: BOUNDS FOR THE EXPONENTIAL FUNCTION

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$.

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Note that we can write $a^x-1=e^{x\log(a)}-1$. Then, using $(1)$ reveals for $x\log(a)<1$

$$x\log(a)\le a^x-1\le \frac{x\log(a)}{1-x\log(a)}\tag 2$$

whereupon applying the squeeze theorem to $(2)$, we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}a^x=1}$$

And we are done!

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$\displaystyle \delta-\epsilon$ PROOF:

To explicitly pursue a $\delta-\epsilon$ proof, we simply break the analysis into cases. For $a>1$, $\log(a)>0$. Next, we choose $|x|<\frac{1}{2\log(a)}$.

Let $\epsilon>0$ be given. Then, from $(2)$, we have for $x>0$

$$\begin{align} |a^x-1|&\le \left|\frac{x\log(a)}{1-x\log(a)}\right|\\\\ &<\epsilon \end{align}$$

when

$$|x|<\delta=\min\left(\frac{1}{2\log(a)},\frac{\epsilon}{(1+\epsilon)\log(a)}\right) \tag 3$$

From $(2)$, we have for $x<0$

$$\begin{align} |a^x-1|&\le \left|x\log(a)\right|\\\\ &<\epsilon \end{align}$$

whenever

$$|x|<\delta=\min\left(\frac{1}{2\log(a)},\frac{\epsilon}{\log(a)}\right) \tag 4$$

Putting together $(3)$ and $(4)$, we have $|a^x-1|<\epsilon$ whenever $|x|<\delta=\min\left(\frac{1}{2\log(a)},\frac{\epsilon}{(1+\epsilon)\log(a)}\right) $

The case for $0<a<1$ is left as an exercise.

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METHODOLOGY $2$: BASIC CALCULUS APPROACH

Now, at the OP's request, we present an approach the avoids use of the natural logarithm function. The approach relies on nothing more than use of the Binomial Series.

Without loss of generality, we analyze the case for which $0<a<1$ and write $a=1-b$ where $0<b<1$. For $a>1$, we simply write $a^x=1/(1/a)^x$ and proceed accordingly).

We write the binomial series for $a^x=(1-b)^x$ as

$$\begin{align} a^x&=\sum_{k=0}^\infty\binom{x}{k}(-b)^k\\\\ &=1+\sum_{k=1}^\infty\binom{x}{k}(-b)^k\\\\ \end{align}$$

For $k\ge 1$ and $|x|<1/2$ it is easy to see that $\left|\binom{x}{k}\right|\le \frac{|x|}{2}$. So, for $|x|<1/2$, we have for any given $\epsilon>0$

$$\begin{align} \left|\sum_{k=1}^\infty\binom{x}{k}(-b)^k\right|&\le |x|\sum_{k=1}^\infty b^k\\\\ &=|x|\frac{b}{1-b}\\\\ &<\epsilon \end{align}$$

whenever $|x|<\delta=\min\left(\frac12,\frac{1-b}{b} \,\epsilon\right)$

as was to be shown!