Basically the above, i cant see how $ g(e^x)=e^x $ using the formal definiton we are saying $ \lim_{h\to 0} \frac{e^{(x+h)}-e^x}h=e^x $
Intuitively i do not see how this can be true.
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7The answer depends on what properties of $e^{x}$ you are allowed to use. – Kavi Rama Murthy Feb 17 '20 at 09:25
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You are allowed to use the formal definition of a limit but not any other limit properties (like L'Hopital) and all properties of e^x except g(e^x)=e^x (obviusly) – Electro Squid Feb 17 '20 at 09:58
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Using exponential properties gets you as far as $e^x\cdot\lim_\limits{h \to 0} \left(\frac{e^{h}-1}{h}\right)$. Which definition of $e$ and $e^x$ are you using/allowed to use - $e^x = \lim_\limits{n \to \infty} \left(1+\frac{x}{n}\right)^n$? – KM101 Feb 17 '20 at 10:20