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February 8th, 2018, 05:09 PM   #11
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I'm talking about the given problem, not the definition of the derivative.
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February 8th, 2018, 05:12 PM   #12
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The given problem is a derivative. Specifically, what is the derivative of $e^x$ at $x = 0$? Well its
\[\lim_{h \to 0} \frac{e^h - 1}{h} \]
which is exactly the limit in the OP.
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February 8th, 2018, 05:17 PM   #13
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That's a specific derivative, The example you gave in your previous post uses the general formula. There's absolutely no reason you can't apply L'Hopital's rule to the limit the OP gave and the result is 1. That result is all I intended to have the OP see. I'm done.
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