Prove that $a^x$ is continuous

Question

I'm having trouble with proving the following:

Let $a > 0$ be a positive real number. Show that the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) := a^x$ is continuous.

I'm a first year mathematics student, so I am trying to prove this in a way that is understandable to me, a delta epsilon proof. Could you help me with this?

Thanks in advance!

Answer 1

For any positive real number a, a^x is defined to mean exp(x ln a) where ln is the inverse of exp. It can be shown that for any 2 functions f and g, if f is continuous on R and g is a linear function with nonzero slope, f ∘ g is continuous so for any positive real number a, if exp(x) is continuous on R, then exp(x ln a) is continuous on R but a^x = exp(x ln a) so a^x is continuous on R if exp(x) is continuous on R. Exp(x) is defined to be the unique function with domain R that both is its own derivative and assigns 1 to 0. It can be shown that a function with that property exists because 1 + x + x²/2! + x³/3! + x⁴/4! ... has that property. It can also be shown that the any function with that property has an inverse with domain (0, ∞) and derivative 1/x over its domain that assigns 0 to 1 but according to https://proofwiki.org/wiki/Zero_Derivative_implies_Constant_Function, all functions that have a 0 derivative in any interval are constant in that interval, so any two functions with domain R that are their own derivative and assign 1 to 0 must have the same inverse and so must be the same function. Therefore, there is exactly one funtion from R to R that's its own derivative and assigns 1 to 0, so we can call it exp(x). Finally, for any x, for any δ in interval (0, exp(x)), we can chose ɛ = δ/2exp(x) and the exp of any real number within (x - ɛ, x + ɛ) will be within (exp(x) - δ, exp(x) + δ) so by definition exp(x) is continuous. Finally, since exp(x) is continuous, a^x is continuous for any positive real number a.

Answer 2

Hey, if you are starting in maths you must use a lot of your imagination, everytime you can, try to draw your problem, I hope it helps you, pick $\delta=\min\{\delta_1,\delta_2\}$

Answer 3

Here's a solution that involves logs; you're going to have to go far out of your way to do so otherwise.

Take $a>1$ WLOG -- if $a<1$, this is just a reflection of an $a>1$ (you should be able to prove that if $f$ is continuous, then $g(x)=f(-x)$ is also continuous pretty easily).

The key is to look for an $x-x_0$ in the function image expression--here that's done by factoring: $a^x-a^{x_0}=a^{x_0}(a^{x-x_0}-1)$. Intuitively now, we know we can make the second element of the product small because we have direct control in choosing $\delta$ over $x-x_0$; the key is to notice that we must also account for the size of $a^{x_0}$ (which is why this function is not uniformly continuous, but that's for another day).

Let's play around with image distance a bit:

$$ \left| a^x-a^{x_0} \right| = \left| a^{x_0} \left( a^{x-x_0} - 1 \right) \right| \leq a^{x_0} \left| a^{x-x_0} - 1 \right| $$

It may take some convincing, but you should prove to yourself that $ |a^y-1| \leq a^{|y|}-1 $ (for starters, look here for at least graphic proof that it works--note that I'm implicitly using that $a>1$).

Thus, $ \left| a^x - a^{x_0} \right| \leq a^{x_0} \left( a^{|x-x_0|} - 1 \right ) $, where we see exactly what we want: $|x-x_0|$.

Now we pick the $\delta$ that will turn the RHS above into just $\varepsilon$--we can do this by setting

$$\delta = \log_a \left( \frac{\varepsilon}{a^{x_0}} + 1 \right) $$

So that:

$\left| a^x-a^{x_0} \right| < \varepsilon $ whenever $|x-x_0|$ is strictly smaller than our chosen $\delta$.

Answer 4

The following is answer without relying on log:

$f(x) := a^x$

$$|a^x - a^{x_0} = a^{x_0}|a^{x - x_0} - 1|$$

Choose $$N \in \mathbb{R}\ : \ \frac{1}{N} \leq x - x_0 \leq \frac{1}{N - 1}$$

It follows:

$$|a^{-N} - 1| \leq |a^{x - x_0} - 1| \leq |a^{-(N + 1)} - 1| $$

As $x \rightarrow x_0$ we have $a^{-N} - 1$ and $a^{-(N + 1)} - 1$ approaches 0.

Therefore by squeeze theorem we have :

$$lim_{x \rightarrow x_0} |a^x - a^x_0| = a^{x_0}|a^{x - x_0} - 1| = 0$$