Question Number 157104 by zakirullah last updated on 19/Oct/21
$${prove}\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\bigtriangleup{x}} −\mathrm{1}}{\bigtriangleup{x}}=\mathrm{1} \\ $$
Answered by puissant last updated on 19/Oct/21
$$\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}\:\sim\underset{\Delta{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+\Delta{x}−\mathrm{1}}{\Delta{x}}=\mathrm{1} \\ $$
Commented by zakirullah last updated on 20/Oct/21
$${thanks}\:{alot} \\ $$
Answered by Fresnel last updated on 20/Oct/21
$$\mathrm{li}\underset{\Delta{x}\dashrightarrow\mathrm{0}} {\mathrm{m}}\frac{{e}^{\Delta{x}} −\mathrm{1}}{\Delta{x}}={li}\underset{\Delta{x}\dashrightarrow\mathrm{0}} {{m}}\frac{{e}^{\Delta{x}} −{e}^{\mathrm{0}} }{\Delta{x}−\mathrm{0}}={e}^{\mathrm{0}} =\mathrm{1}\:{becausef}'\left({x}\right)={e}^{{x}\:} {for}\:{f}\left({x}\right)={e}^{{x}} \:{and}\:{li}\underset{{x}\dashrightarrow\mathrm{0}} {{m}}\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}={f}'\left(\mathrm{0}\right). \\ $$
Commented by zakirullah last updated on 20/Oct/21
$${great}\:{sir}! \\ $$