Antwort
$\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.
Erklärung
Also $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ and $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$.
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What is the standard limit $\lim_{x \to 0} \frac{\sin x}{x}$ and $\lim_{x \to 0} \frac{e^x - 1}{x}$?
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