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

  • $\begingroup$ If you understand the derivation of $f(x)=e^x$, then simply write $h=x\ln(a)$ and $e^{x_0}=\ln(a)$. $\endgroup$ – Demosthene May 2 '15 at 13:45
  • $\begingroup$ Would this be of any help ? $\endgroup$ – Lucian May 2 '15 at 14:42

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 $x \to 0$.


Put $\;f(x)=a^x\;$ , then



$$f'(x)=a^x\log a\implies f'(0)=a^0\log a=\log a$$


Just remark that

$$\frac{e^{x\ln(a)}-1}{x \ln(a)}$$

Is of the form



$$\lim_{x\to 0}\frac{e^{x\ln(a)}-1}{x \ln(a)} = \left( x\mapsto e^{x\ln(a)}\right)'(0) = \ln(a) e^{0 \ln(a)} = \ln(a)$$


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 $L=\ln a/\ln e=\ln a$.


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