EDİTED: Find the derivative of $f(x)=a^x$, using the definition of the derivative. EDİTED:

I want to ask MSE to confirm the correctness of the alternate solution and its mistake.

My previous attempt was wrong,That's why I made a new initiative.
Using the definition of the derivative I wanted to find the derivative of the function $f(x)=a^x$.
Here $a≠0,a\in \mathbb{R^{+}}$ and $a^x=e^{x \ln a}$
I wrote these:
$(a^x)'=\lim_{\delta x\to 0}\frac{a^{x+\delta x}-a^x}{\delta x}=a^x×\lim_{\delta x\to 0}\frac {a^{\delta x}-1}{\delta x}$

Now,I must find $\lim_{\delta x\to 0}\frac {a^{\delta x}-1}{\delta x}$

I tried to do something:

$$a^{\frac{\delta x}{a^{\delta x}-1}}=\left(1+\left(a^{\delta x}-1\right)\right)^{\frac{1}{a^{\delta x}-1}}$$
$$\lim_{\delta x \to 0} {a^{{\frac {\delta x}{a^{\delta x}-1}}}}=\lim_{\delta x \to 0} {{\left(1+\left(a^{\delta x}-1\right)\right)^{\frac{1}{a^{\delta x}-1}}}}$$
$$a^{\lim_{\delta x\to 0}{{{\frac {\delta x}{a^{\delta x}-1}}}}}=\lim_{\delta x \to 0} {{\left(1+\left(a^{\delta x}-1\right)\right)^{\frac{1}{a^{\delta x}-1}}}}$$
$$a^{\lim_{\delta x\to 0}{{{\frac {\delta x}{a^{\delta x}-1}}}}}=e$$
$$\lim_{\delta x\to 0}{{{\frac {\delta x}{a^{\delta x}-1}}}}=\log_a{e}$$
$$\frac{1}{\lim_{\delta x\to 0}{{{\frac {\delta x}{a^{\delta x}-1}}}}}=\frac{1}{\log_a{e}},\log_a{e}≠0$$
$$\lim_{\delta x\to 0} \frac {1}{ \frac{\delta x}{a^{\delta x}-1}}=\ln a$$
$$\lim_{\delta x\to 0}\frac {a^{\delta x}-1}{\delta x}=\ln a$$

I used:


*

*$\lim_{n\to 0}{n}=\lim_{\delta x\to 0}{(a^{\delta x}-1)}=a^0-1=0$


*$\lim_{\delta x \to 0} {{\left(1+\left(a^{\delta x}-1\right)\right)^{\frac{1}{a^{\delta x}-1}}}}=\lim_{n\to 0}{(1+n)^{\frac 1n}}=e$

Finally,

$$(a^x)'=\lim_{\delta x\to 0}\frac{a^{x+\delta x}-a^x}{\delta x}=a^x×\lim_{\delta x\to 0}\frac {a^{\delta x}-1}{\delta x}=a^x \ln a$$

Is this method/way/solution correct?
Thank you!
 A: There are some subtleties here. 


*

*How is $a^x$ defined, for a generic $a>0$ and some $x\in\mathbb{R}$? The most common ways are to define $a^x$ directly as $\exp\left(x\log a\right)$, or to consider a sequence of rational numbers $\left\{\frac{p_n}{q_n}\right\}_{n\geq 0}$ convergent to $x$ and let $a^x=\lim_{n\to +\infty}a^{\frac{p_n}{q_n}}$ ;

*If we have $\lim_{n\to +\infty}f(n)=L$ (limit of a sequence) it is not granted that $\lim_{x\to +\infty} f(x)=L$ (limit of a function) without further assumptions on $f$. For instance $\lim_{n\to +\infty}\sin(\pi n)=0$ but $\lim_{x\to +\infty}\sin(\pi x)$ does not exist, so we have to be careful in deriving $\lim_{x\to 0}\frac{a^x-1}{x}=\log a$ from $\lim_{n\to +\infty}\frac{a^{1/n}-1}{1/n}=\log a$;

*On the other hand, there is no need to over-complicate things: $\lim_{x\to 0}\frac{a^x-1}{x}$ is just the value of the derivative of $a^x$ at the origin. If we know/prove that $a^x=\exp\left(x\log a\right)$, then $\lim_{x\to 0}\frac{a^x-1}{x}=\log a$ is a straightforward consequence of the chain rule, given the differentiability of the exponential function.
A: User Jack D'Aurizio has raised some good points in his answer and I want to add some remarks in that direction. So this is not exactly an answer but more of a comment which does not fit in comment box.

For most students the first encounter with exponential and logarithmic functions is not a very fascinating one. They are presented with lots of nice (and even strange) properties of these functions and usually the definitions and proofs are absent. Armed with this information the student simply can not evaluate limits as the one in question and needless to say the proof / solution is incorrect.
Textbook authors should be more forthcoming and intellectually honest to say that a proper theory of these functions can not be provided at this stage and the properties of these functions have to be assumed (translate: mugged up) without any proof whatsoever. And they should not give exercises to prove these properties rather the exercises should use these assumed properties to handle more challenging problems. Thus the current question is pointless unless the proper theory of these functions has been developed.
Instead of asking "Evaluate $\lim\limits_{x\to 0}\dfrac{a^x-1}{x}$" it is better to give the following exercise :

Evaluate the limit $\lim\limits_{x\to 0}\dfrac{2^x+3^x-2}{x}$. You may use the limit formula $\lim\limits_{x\to 0}\dfrac{a^x-1}{x}=\log a$ and properties of logarithms.


Update: I decided to answer your question in simple English. Your approach has the following issues:


*

*no definition of $a^x$ is provided.

*you need to show that $\lim\limits_{\delta x\to 0}a^{\delta x} =1$

*you need to define $e$ and show that $\lim\limits_{\delta x\to 0}(1+\delta x) ^{1/\delta x} = e$. 

*you need to also define $\log a$.


Fixing all the above issues is difficult. Some of these problems have been fixed. See below.
Further Update: the question has been edited to include some definitions (like for $a^x$) and some assumptions about $e$ and logarithmic function have been used. With these assumptions one can say that your approach is correct.
Also OP has raised a concern in comments: does one need to prove each and every theorem being used in order to solve a problem? NO!! But only when the theorem is well known and supposed to be considered basic. If one is trying to prove the basic results then one has to be extra cautious and mention the definitions and prove the basic results using those definitions. For example, one can use L'Hospital's Rule to evaluate a limit, but when one has to prove L'Hospital's Rule, it is necessary to go deeper and start with mean value theorems.
A: 
I thought it might be instructive to present a way forward the relies on pre-calculus tools only.  To that end we proeed.


If we define $a^x$ by the expression $e^{\log(a)\,x}$, then we can write
$$\frac{a^x-1}{x}=\frac{e^{\log(a)\,x}-1}{x}\tag 1$$


In THIS ANSWER, I used only the limit definition of the exponential function along with Bernoulli's Inequality to show the exponential function satisfies the inequalities
$$1+x\le e^x\le \frac{1}{1-x}\tag2$$
for $x<1$.


Applying $(2)$ to $(1)$ reveals for $0<x<1$
$$ \log(a) \le \frac{a^x-1}{x}\le \frac{\log(a)}{1-\log(a) x}\tag3$$
and for $x<0$
$$\frac{\log(a)}{1-\log(a)x}\le \frac{a^x-1}{x}\le \log(a)\tag 4$$
Finally, applying the squeeze theorem to $(3)$ and $(4)$ yields the coveted limit

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

And we are done!
A: I guess one way to proceed is as follows. I use $\Delta$ to simplify notation. You need $$\lim_{\Delta \to 0} \frac{a^{\Delta} - 1}{\Delta}$$. Let $a = e^b$ or $b = \ln(a)$. multiplying numerator and denominator by $b$, you get the above equals $$b \cdot \lim_{\Delta \to 0} \frac{e^{b\Delta} -1}{b\Delta}$$ which is the derivative of $e^x$ at $x=0$ which is 1. So the expression you want is $b = \ln a$.
A: So here is a quick idea: let $\delta x = \log_a(1 + \frac{1}{n}) \to 0$
$$ \lim_{\delta x \to 0} \frac{a^{\delta x} - 1}{\delta x} = \lim_{n \to \infty} \frac{1}{n\cdot \log_a(1 + \frac{1}{n})} = \lim_{n \to \infty} \left(\log_a(1 + \frac{1}{n})^n \right)^{-1} \to \log_a(e)^{-1} = \ln(a)$$ 
A: If we accept
$$e^{x} = \lim_{n \rightarrow \infty} \left(1 + \frac{x}{n}\right)^{n}$$
Then we try to prove:
\begin{equation} \lim_{x \rightarrow 0} \frac{e^{x} - 1}{x} = 1 \end{equation}
This will solve our problem, because then:
$$\lim_{h \rightarrow 0}\frac{a^{h} - a}{h} = \lim_{h \rightarrow 0}\frac{e^{h\ln(a)} - 1}{h} = \ln(a)\lim_{t \rightarrow 0}\frac{e^{t} - 1}{t} = \ln(a)$$
Where we use the substitution $t = h\ln(a)$. Now use the binomial theorem:
$$\lim_{x \rightarrow 0}\frac{e^{x} - 1}{x} = \lim_{x \rightarrow 0}\frac{\lim_{n \rightarrow \infty}\sum_{k = 0}^{n}\binom{n}{k}\frac{x^{k}}{n^{k}} - 1}{x} = 1 + \lim_{x \rightarrow 0}\lim_{n \rightarrow \infty}\sum_{k = 2}^{n}\binom{n}{k}\frac{x^{k-1}}{n^{k}}$$
Since $\binom{n}{k} \leq n^{k}$ for all $k \geq 0$, we have:
$$\left|\sum_{k = 2}^{n}\binom{n}{k}\frac{x^{k-1}}{n^{k}}\right| \leq \sum_{k = 1}^{n-1} |x|^{k} \leq \sum_{k=1}^{\infty}|x|^{k} = \frac{|x|}{1 - |x|}$$
And so:
$$\lim_{x \rightarrow 0}\lim_{n \rightarrow \infty} \sum_{k = 2}^{n}\binom{n}{k}\frac{x^{k-1}}{n^{k}} = 0$$
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