Respuesta
$\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.
Explicación
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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Flashcard de estudio del mazo JEE Main Master Class: Calculus, Mechanics, Coordinate Geometry.
$\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$.
Also $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ and $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$.