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

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The limit of (a^x - 1)/x as x approaches 0 is log(a). This result can be derived using L'Hospital's rule by differentiating the numerator and denominator. Alternatively, the Maclaurin series expansion of a^x can be used to simplify the expression, leading to the same conclusion. Both methods confirm that the limit equals log(a). Understanding these approaches provides insight into evaluating limits involving exponential functions.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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