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

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

###### Corollary.

For $a>0$, we have

$\lim_{{x\to 0}}\frac{a^{x}-1}{x}=\ln a.$ |

###### Proof.

Recall that $a^{x}=e^{{x\ln a}}$. Thus,

$\displaystyle\lim_{{x\to 0}}\frac{a^{x}-1}{x}$ | $\displaystyle=\lim_{{x\to 0}}\frac{e^{{x\ln a}}-1}{x}$ | ||

$\displaystyle=\lim_{{x\to 0}}\frac{(e^{{x\ln a}}-1)\ln a}{x\ln a}$ | |||

$\displaystyle=(\ln a)\lim_{{x\to 0}}\frac{e^{{x\ln a}}-1}{x\ln a}.$ |

Let $t=x\ln a$. Then $t\to 0$ as $x\to 0$. Therefore,

$\displaystyle\lim_{{x\to 0}}\frac{a^{x}-1}{x}$ | $\displaystyle=(\ln a)\lim_{{t\to 0}}\frac{e^{t}-1}{t}$ | ||

$\displaystyle=(\ln a)1$ | |||

$\displaystyle=\ln a.\qed$ |

The formula from the corollary is useful for proving that $\displaystyle\frac{d}{dx}a^{x}=a^{x}\ln a$. On the other hand, once this fact is known, the corollary is easily proven via l’Hôpital’s rule:

$\displaystyle\lim_{{x\to 0}}\frac{a^{x}-1}{x}$ | $\displaystyle=\lim_{{x\to 0}}\frac{a^{x}\ln a}{1}$ | ||

$\displaystyle=a^{0}\ln a$ | |||

$\displaystyle=\ln a.$ |

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## Comments

## lim(x->0) (a^x - 1) / x?

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?

## Re: lim(x->0) (a^x - 1) / x?

>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