Having introduced the basic definitions and properties of exponential functions, we now introduce some proper calculus notions into our analysis, starting with continuity. As with algebraic functions, it turns out that exponential functions are continuous everywhere on their domains. Although we are not in a position to give a formal proof of this fact, continuity of these functions is plainly evident from the graphs we have examined so far, which have no breaks or holes.
Continuing with our limit analysis, we next examine the βend behaviorβ of an exponential function
\(f\text{:}\) that is, we wish to compute
\(\lim\limits_{x\to \pm \infty}f(x)\text{.}\)
To start, letβs consider the two basic exponential functions
\(f(t) = 2^t\) and
\(g(t) = \left(\frac{1}{2}\right)^t\) and their respective values at
\(t=10, t=20\text{,}\) and
\(t=30\text{.}\)
For the increasing function \(f(t) = 2^t\text{,}\) we see that the output of the function gets very large very quickly. In addition, there is no upper bound to how large the function can be. Indeed, we can make the value of \(f(t)\) as large as weβd like by taking \(t\) sufficiently big. We thus say that as \(t\) increases, \(f(t)\) increases without bound and conclude
\begin{equation*}
\lim_{t\to\infty}2^t = \infty
\end{equation*}
For the decreasing function \(g(t) = \left(\frac{1}{2}\right)^t\text{,}\) we see that the output \(g(t)\) is always positive but getting closer and closer to 0. Indeed, because we can make \(2^t\) as large as we like, it follows that we can make its reciprocal \(\frac{1}{2^t} = \left(\frac{1}{2}\right)^t\) as small as weβd like. We thus say that as \(t\) increases, \(g(t)\) approaches 0: i.e.,
\begin{equation*}
\lim_{t\to\infty}\left(\frac{1}{2}\right)^t = 0\text{.}
\end{equation*}
What about the limits of these functions at \(-\infty\text{?}\) It turns out these limits are essentially equal to the reciprocal of the limits at \(\infty\text{,}\) thanks to the exponential property \(b^{-x}=1/b^x\text{.}\) Assuming our results from above, we have
\begin{align}
\lim\limits_{x\to -\infty} 2^x \amp = \lim\limits_{x\to \infty}2^{-x} \amp (\lim\limits_{x\to -\infty}f(x)=\lim\limits_{x\to \infty}f(-x)) \tag{2.9}\\
\amp = \lim\limits_{x\to \infty}(1/2)^x\tag{2.10}\\
\amp = 0\tag{2.11}\\
\lim\limits_{x\to -\infty}(1/2)^x \amp = \lim\limits_{x\to \infty}(1/2)^{-x} \amp (\lim\limits_{x\to -\infty}f(x)=\lim\limits_{x\to \infty}f(-x))\tag{2.12}\\
\amp = \lim\limits_{x\to \infty}2^x \tag{2.13}\\
\amp = \infty\text{.}\tag{2.14}
\end{align}
Our analysis above, examining the special cases \(b=2\) and \(b=1/2\) generalizes easily to arbitrary positive bases \(b\text{.}\) The crucial feature of \(b\) is whether it is greater than or less than \(1\) as the following theorem articulates.
We have established (or at least stated) two important results about exponential functions.
-
Exponential functions
\(f(x)=b^x\) are continuous on their domain
\(\R\text{;}\)
-
One limit at infinity of an exponential function
\(f(x)=b^x\) is
\(\infty\text{,}\) and the other is
\(0\text{.}\)
From these two facts, we can deduce a third important fact about the range of an exponential theorem, using our old friend, the
intermediate value theorem: namely,
\(\range f=(0,\infty)\) for any exponential function
\(f(x)=b^x\text{.}\)
Letβs see why. Assume
\(f(x)=b^x\) with
\(b> 1\text{.}\) (The case
\(0< b< 1\) is exactly similar.) Given any positive
\(d\in \R\text{,}\) since
\(\lim\limits_{x\to \infty}f(x)=\infty\text{,}\) we can find a positive
\(b\in \R\) such that
\(f(b)> c\text{;}\) similarly, since
\(\lim\limits_{x\to -\infty}f(x)=0\) we can find a negative
\(a\in \R\) such that
\(f(a)< d\text{.}\) We thus have
\(a< b\text{,}\) \(f\) continuous on
\([a,b]\) (since
\(f\) is continuous everywhere), and
\(f(a)< d < f(b)\text{.}\) We conclude that there exists some
\(c\in (a,b)\) such that
\(f(c)=d\text{.}\) The argument shows that for any positive
\(d\in (0,\infty)\text{,}\) we have
\(d\in \range g\text{.}\) Since furthermore
\(\range f\subseteq (0,\infty)\text{,}\) we conclude that
\(\range f=(0,\infty)\text{.}\)