Show that exponentiation is continuous. Proposition: Let $a>0$ be a positive real number. Then the function $f:\Bbb{R}\rightarrow\Bbb{R}$ defined $f(x):=a^x$ is continuous. So far what I got seems really far-off from what I think the text is asking.
Proof Attempt:
We want to show that $\forall\epsilon>0\exists\delta>0$ such that if $\vert x-x_0 \vert<\delta$, then $\vert a^x-a^{x_0}\vert<\epsilon$. Computing for $\delta$ 
\begin{align}
\vert a^x-a^{x_0}\vert<\epsilon &\Longleftrightarrow a^{x_0}-\epsilon<a^x<a^{x_0}+\epsilon\\
&\Longleftrightarrow \log_a(a^{x_0}-\epsilon)<\log_a(a^x)<\log_a(a^{x_0}+\epsilon)\\ &\Longleftrightarrow \log_a(a^{x_0}-\epsilon)<x<\log_a(a^{x_0}+\epsilon)\\
\end{align}
Hence, let $\delta=\min(\log_a(a^{x_0}-\epsilon), \log_a(a^{x_0}+\epsilon))$. QED.
Is this correct? Since I am working through Tao's Analysis text he discourages the reader from using logarithms since logarithms wasn't defined on the text. Instead, he hinted that we use the following propositions(I want to know how to use these propositions):

Lemma 6.5.3. For any $x>0$, we have
  $lim_{n\rightarrow\infty}x^{1/n}=1$.
Corollary 6.4.14. Let $(a_n)_{n=m}^{\infty}$, $(b_n)_{n=m}^{\infty}$,
  and $(c_n)_{n=m}^{\infty}$ be sequences of real numbers such that
  $a_n\leq b_n\leq c_n$ for all $n\geq m$. Suppose also that
  $(a_n)_{n=0}^{\infty}$ and $(c_n)_{n=0}^{\infty}$ converges to $L$.
  Then $(b_n)_{n=0}^{\infty}$ is also convergent to $L$.
  

 A: I would say it is not correct, since defining log (the inverse of exponentiation) does not make so much sense until you already know that exponentiation is a continuous bijection onto $(0,\infty)$.  
I assume $a^x$ for real $x$ is defined as a limit of $a^x$ with rational $x$?  Assuming it has been discussed, I would use the limit definition of continuity, showing that if you have a sequence approaching $x$ then applying $f$ gives a sequence approaching $a^x$.  Try to bound this sequence on both sides using rational sequences (for which you know the limit by definition) and then apply Corollary 6.4.14
A: I think you can extend Lemma 6.5.3 to the following result,
Let $(a_n)_{n \in \mathbb N}\subseteq \mathbb R$ such that $\lim_{n\to\infty} a_n= a$. Then $\lim_{n\to\infty} x^{a_n-a}=1$, for every $x>0$.
Therefore, showing that $f=x \to a^x$ is continous is equivalent to show that for every $(x_n)_{n\in \mathbb N}$ converging to $x \in \mathbb R$, we have $\lim_{n\to\infty} a^{x_n}=a^x$. Indeed,
$$
\left|a^{x_n}-a^x \right|=a^{x}\left|a^{x_n-x}-1 \right|\to 0 \quad \text{if} \quad x_n\to x.
$$
A: Well this is how one can sort this out. First and foremost your lemma 6.5.3 is simple and important here.
Using the lemma one can prove that $\lim_{x\to 0}a^x=1$ and then using the power rule $a^{x+h} =a^xa^h$ one proves the continuity of $f(x) =a^x$.
So the key is to prove $a^x\to 1$ as $x\to 0$. To that end we can start with $\epsilon >0$ and using lemma 6.5.3 there is a positive integer $m$ such that $$|a^{1/n}-1|<\epsilon >|a^{-1/n}-1|$$ whenever $n\geq m$. Let $\delta=1/m$ and then if $0<|x|<\delta$ then $a^x$ lies between $a^{1/m}$ and $a^{-1/m}$ and hence between $1-\epsilon$ and $1+\epsilon$. It now follows that $a^x\to 1$.
