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Question Number 140207 by mathlove last updated on 05/May/21

Answered by liberty last updated on 06/May/21

$$\underset{△x\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}+\mathrm{m}}.{\mathrm{e}}^{△\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}+\mathrm{m}}}{△\mathrm{x}}=$$$${\mathrm{e}}^{\mathrm{x}+\mathrm{m}}.\underset{△x\to 0}{\lim }\frac{{\mathrm{e}}^{△\mathrm{x}}-1}{△\mathrm{x}}=$$$${\mathrm{e}}^{\mathrm{x}+\mathrm{m}}.\underset{△x\to 0}{\lim }\frac{{\mathrm{e}}^{△\mathrm{x}}}{1}={\mathrm{e}}^{\mathrm{x}+\mathrm{m}}.$$

Answered by wassim last updated on 05/May/21

$$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{e^{(x+m)+\mathrm{\Delta }x}-e^{x+m}}{\mathrm{\Delta }x}=\frac{d}{d(x+m)}{e}^{x+m}=e^{x+m}$$

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