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Question Number 157104 by zakirullah last updated on 19/Oct/21

$$prove\underset{△x\to 0}{\lim }\frac{e^{△x}-1}{△x}=1$$

Answered by puissant last updated on 19/Oct/21

$$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{e^{\mathrm{\Delta }x}-1}{\mathrm{\Delta }x}\sim \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{1+\mathrm{\Delta }x-1}{\mathrm{\Delta }x}=1$$

Commented by zakirullah last updated on 20/Oct/21

$$thanksalot$$

Answered by Fresnel last updated on 20/Oct/21

$$\mathrm{li}\underset{\mathrm{\Delta }x⇢0}{\mathrm{m}}\frac{e^{\mathrm{\Delta }x}-1}{\mathrm{\Delta }x}=li\underset{\mathrm{\Delta }x⇢0}{m}\frac{e^{\mathrm{\Delta }x}-e^0}{\mathrm{\Delta }x-0}=e^0=1{becausef}^{\prime }\left(x\right)=e^{x}forf\left(x\right)=e^{x}andli\underset{x⇢0}{m}\frac{f\left(x\right)-f\left(0\right)}{x-0}=f^{\prime }\left(0\right).$$

Commented by zakirullah last updated on 20/Oct/21

$$greatsir!$$

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