$\lim _{x\to 0}\left(\frac{a^x-1}{x}\right)$ I replace ${a^x-1} = t$ but dont know, what doing next.
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Apply L'Hospital immediatly. – Cornman Oct 01 '17 at 12:08
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1need without this rule. – Dmitry Sokolov Oct 01 '17 at 12:10
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@Cornman How do you take the derivative of $a^x$? – Simply Beautiful Art Oct 01 '17 at 12:11
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@DmitrySokolov How do you define $a^x$? – Simply Beautiful Art Oct 01 '17 at 12:11
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@SimplyBeautifulArt: $a^x=e^{\ln(a)x}$ – Cornman Oct 01 '17 at 12:19
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1@Cornman Yes, and how do you take the derivative of $e^{\ln(a)x}$? The point is that the derivative is defined to be the limit in question, so L'Hospital's sends us straight to the original problem. – Simply Beautiful Art Oct 01 '17 at 12:21
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Oh, thank you. I did not think about that. – Cornman Oct 01 '17 at 12:22
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This is the definition of the derivative of the function $y \mapsto a^y$ at $y=0$. Equivalently, $y \mapsto \exp(y \log(a))$, which you can differentiate using the chain rule.
Patrick Stevens
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