What is the limit of (a^x - 1)/x as x approaches 0?

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Homework Statement



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

Homework Equations



NA

The Attempt at a Solution



The professor told me that the answer to that limit is log(a), but why? I don't understand; can someone explain why?
 
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Do you know l'hopital's rule?
 
gb7nash said:
Do you know l'hopital's rule?


Yeah, I just figured it out. Take the derivative of both the numerator and the denominator and then it's easy from there...

Is there any way to do it without having to use L'Hospital's?
 
You could take the maclaurin series of a^x and plug it in for a^x. Once you do this, stuff cancels out and you'll obtain the same answer.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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