Proving that the exponential function is continious at $0$

Question

Given $b \in \mathbb R,b>0$, I have the exponential function $f(x)=b^x$ and I want to show that it is continious at $x=0$. From epsilon-delta definition, given an $\epsilon>0$, I need to find $\delta$ s.t $$|x-0|<\delta \implies |b^x-b^0|<\epsilon \text{, or}$$ $$|x|<\delta \implies |b^x-1|<\epsilon.(1)$$

From $(1)$, I have $$1-\epsilon<b^x<1+\epsilon$$ and by taking the logarithm of both sides in base b, I get

$$log_b(1-\epsilon)<x<log_b(1+\epsilon).$$

So I have chosen $\delta=min\text{{$log_b(1+\epsilon),-log_b(1-\epsilon)$}}.$

Is this proof valid? If not, can you give me a valid proof using the epsilon-delta definition or the sequential characterization of continiuty?

Answer 1

The function is convex

$\lambda f(a) + (1-\lambda) f(b) \ge f(\lambda a + (1-\lambda)b)$ with $0\le\lambda\le 1$

Or more relevant to this problem. $\lambda f(1) + (1-\lambda) f(0) \ge f(\lambda)$

But, it seems to me we need two cases for when $b>1$ and when $0<b\le 1$

Let $\delta = \min(1,\frac {\epsilon}{|f(1)|},\frac {\epsilon}{|f(-1)|})$

Answer 2

Speaking about the squential characterization: take a sequence $x_n$ that converges to zero. I want to prove that $b^{x_n}\rightarrow 1$. Since $x_n$ is infinitesimal, for every $\varepsilon >0$ you can say that from a certain $n$ on, you have $|x_n|<\varepsilon$, in other words $-\varepsilon<x_n<\varepsilon$ now you take the exponential and you are done (just observe that if $0<b<1$ you need to switch the $b^\varepsilon$ and $b^{-\varepsilon}.$)