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limit of $\displaystyle \frac{a^x-1}{x}$ as $x$ approaches 0

Major Section: 
Reference
Type of Math Object: 
Corollary

Mathematics Subject Classification

32A05 no label found

Comments

Isn't the proof circular? This limit is normally needed to investigate the exponential function a^x, before the number e has been defined; therefore ln(a) cannot be used.
I think that a more logical (or pedagogical?) approach would be to start computing the derivative of a^x directly from the definition. It is found to be A (a^x), where A is just the limit above; it is a function L(a) of a. For another base b, it is easy to show that L(b) = A log_a (b). Therefore the functional form of L(a) is known up to a constant. Now, we can define e such that L(e) = 1 or:
lim(x->0) (e^x - 1) / x = 1. What would be the link between this definition and the traditional one?

>Isn't the proof circular?
I don't believe Danny. However I think isn't necessary circumvent the proof in such a way and on that point you're right, in my opinion. As one may start directly from the definition about the derivative of a^x, i.e.
(a^x)'=a^x\lim_{y\to x}[e^{(y-x)\log a}-1]/(y-x),
and once we have proved the derivative on polynomials, will be easy to get Taylor through f(x)=\sum_{n=0}^\infty c_n x^n, but all this last part may be omitted. From this we expand out e^u and so
e^{(y-x)\log a}-1= (y-x)\log a/1!+...,
readily obtaining {a^x)'=a^x\log a, as you have bounded. From here, the remaining is clear. Obviously, another ways are possible.
perucho

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