Troubles understanding this limit I have troubles understanding this limit:
$$\lim_{x\to0} \frac{a^x -1}{x}=ln( a)$$
I have the following proof:
$$\frac{a^x -1}{x}=\frac{e^{xlna}-1}{x}=\frac{e^{xlna}-1}{x ln(a)}ln(a) \xrightarrow{x\to0} ln(a)$$
Is there a way to understand this without the use of the series of the exponential function and without L'Hôpital? I can see by plotting the function how the function behaves, but is there an analytical way to understand this? Do I miss a trick or something?
Why is this term:$$\frac{e^{xlna}-1}{x ln(a)}$$
going to $1$ instead of $0$? 
A similar example would be this:
Let $f(x)=e^x$. The derivation of $x_0$ is given through:
$$f'(x_0)=\lim_{h\to0}\frac{e^{x_0+h}-e^{x_0}}{h}=e^{x_0}\lim_{h\to0}\frac{e^{h}-1}{h}=e^{x_0}$$
Thank you
 A: A proper answer to your question requires that you know the proper definitions of $e^{x}, \log x, a^{x}$. Unfortunately a sound theory of these functions is not provided in beginner's calculus texts (see my blog series for an exposition of these topics).
If you don't want to work through a sound theory of these functions then you will have to rely on certain assumptions without proof. Thus we have two assumptions here
1) $a^{x} = e^{x \log a}$ (this is one of the accepted definitions in a sound theory of these functions mentioned above).
2) $\lim\limits_{x \to 0}\dfrac{e^{x} - 1}{x} = 1$.
Then since $x \to 0$ it follows that $y = x \log a \to 0$ and therefore $\dfrac{e^{y} - 1}{y} \to 1$ i.e. $$\frac{e^{x \log a} - 1}{x\log a} \to 1$$ as $x \to 0$.
A: Put $\;f(x)=a^x\;$ , then
$$f'(0):=\lim_{x\to0}\frac{f(x)-f(0)}x=\lim_{x\to0}\frac{a^x-1}x$$
But
$$f'(x)=a^x\log a\implies f'(0)=a^0\log a=\log a$$
A: Just remark that 
$$\frac{e^{x\ln(a)}-1}{x \ln(a)}$$
Is of the form 
$$\frac{f(h)-f(0)}{h-0}$$
Hence 
$$\lim_{x\to 0}\frac{e^{x\ln(a)}-1}{x \ln(a)} = \left( x\mapsto e^{x\ln(a)}\right)'(0) = \ln(a) e^{0 \ln(a)} = \ln(a)$$
A: I have a satisfying proof.
We calculate the limit $L=\lim_{x\to 0}\frac{a^x-1}{x}$ using the substitution method. 
Let $a^x-1=y$ so if $x\to 0 \Rightarrow a^x-1\to 0 \Rightarrow  y\to 0$ and $x=\log_a(y+1)$
We rewrite the limit as 
$$L=\lim_{y\to 0}\ \frac{y}{\log_a (y+1)}$$
But from the log formula $\log_a(y+1)=\frac{\ln(y+1)}{\ln(a)}$ and the limit becomes 
$$L=lim_{y\to 0}\frac{y\ln a}{\ln(y+1)} =\ln (a) \lim_{y\to 0}\frac{1}{(1/y)*ln(y+1)}=\ln(a) \lim_{y\to 0}\frac{1}{\ln (y+1)^{1/y}}$$
But we know the definition of the number $e(=2.718..)=\lim_{x\to 0}(1+x)^{1/x}$
Being the base of the natural logarithm, the limit becomes $L=\ln a/\ln e=\ln a$.
