Question Number 219831 by SdC355 last updated on 02/May/25
$$\mathrm{prove} \\ $$$$\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{g}\left({z}+{h}\right)}{\mathrm{g}\left({z}\right)}\right)^{\frac{\mathrm{1}}{{h}}} ={e}^{\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}\:\mathrm{ln}\:\left(\mathrm{g}\left({z}\right)\right)} ={e}^{\frac{\mathrm{g}^{\left(\mathrm{1}\right)} \left({z}\right)}{\mathrm{g}\left({z}\right)}} \\ $$
Answered by MrGaster last updated on 04/May/25
$$\mathrm{lim}_{{h}\rightarrow\mathrm{0}} \left(\frac{\mathrm{g}\left({z}+{h}\right)}{\mathrm{g}\left({z}\right)}\right)^{\frac{\mathrm{1}}{{h}}} =\mathrm{exp}\left(\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{ln}\left({g}\left({z}+{h}\right)\right)−\mathrm{ln}\left({g}\left({z}\right)\right)}{{h}}\right)=\mathrm{exp}\left(\frac{{d}}{{dz}}\mathrm{ln}\left({g}\left({z}\right)\right)\right)={e}^{\frac{{g}'\left({z}\right)}{{g}\left({z}\right)}} \\ $$