\[ \lim_{x \to 0} \frac{e^x - e^{-x}}{x} = \frac{e^0 - e^{0}}{0} = \frac{1 - 1}{0} \] Aplicando a regra de L'Hôpital: \[ \lim_{x \to 0} \frac{e^x - e^{-x}}{x} = \lim_{x \to 0} \frac{e^x + e^{-x}}{1} = 2 \]

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on Oct 25, 2024 at 2:06 pm



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