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Question Number 180413 by Mastermind last updated on 12/Nov/22

$$\mathrm{li}\underset{\mathrm{h}\to 0}{\mathrm{m}}\left(\frac{{\mathrm{e}}^{\mathrm{h}}-1}{\mathrm{h}}\right)$$$$$$$$\mathrm{find}\mathrm{the}\mathrm{limit}\mathrm{above}$$

Answered by Frix last updated on 12/Nov/22

$$=\underset{x\to 0}{\lim }\frac{{\mathrm{e}}^x}{1}=1\left({\mathrm{l}}^{\prime }\mathrm{H}\widehat{\mathrm{o}pital}\right)$$

Answered by mr W last updated on 12/Nov/22

$$=\underset{h\to 0}{\lim }\frac{e^h-e^0}{h}$$$$=\left(\frac{{de}^x}{dx}\right){\mid }_{x=0}$$$$=\left(e^x\right){\mid }_{x=0}$$$$=1$$

Commented by Frix last updated on 12/Nov/22

$$\mathrm{yes}!$$

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