lim-h-0-e-h-1-h-find-the-limit-above-

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

li \underset{h \to 0}{m} (\frac{e^{h} - 1}{h})

find the limit above

Answered by Frix last updated on 12/Nov/22

= \lim_{x \to 0} \frac{e^{x}}{1} = 1 (l^{'} H \hat{o p i t a l})

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

= \lim_{h \to 0} \frac{e^{h} - e^{0}}{h}

= (\frac{{d e}^{x}}{d x}) ∣_{x = 0}

= (e^{x}) ∣_{x = 0}

= 1

Commented by Frix last updated on 12/Nov/22

yes!

Facebook Tweet Pin