Question Number 180413 by Mastermind last updated on 12/Nov/22
$$\mathrm{li}\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{m}}\left(\frac{\mathrm{e}^{\mathrm{h}} −\mathrm{1}}{\mathrm{h}}\right) \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{above} \\ $$
Answered by Frix last updated on 12/Nov/22
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{e}^{{x}} }{\mathrm{1}}\:=\mathrm{1}\:\:\:\:\:\left(\mathrm{l}'\mathrm{H}\hat {\mathrm{o}pital}\right) \\ $$
Answered by mr W last updated on 12/Nov/22
$$=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{h}} −{e}^{\mathrm{0}} }{{h}} \\ $$$$=\left(\frac{{de}^{{x}} }{{dx}}\right)\mid_{{x}=\mathrm{0}} \\ $$$$=\left({e}^{{x}} \right)\mid_{{x}=\mathrm{0}} \\ $$$$=\mathrm{1} \\ $$
Commented by Frix last updated on 12/Nov/22
$$\mathrm{yes}! \\ $$