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2.8: Límites algebraicos
https://espanol.libretexts.org/Matematicas/Calculo_informal_con_aplicaciones_a_las_ciencias_biologicas_y_ambientales_(Seacrest)/02%3A_Introducci%C3%B3n_Derivada/2.08%3A_L%C3%ADmites_algebraicos
\[\begin{align*} \lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3} &= \lim_{x \to 3} \frac{(x - 3)(x + 1)}{x - 3} \\ & = \lim_{x \to 3} \frac{\cancel{(x - 3)}(x+1)}{\cancel{x-3}} \\ & = \lim_{x \to 3} x + 1 \... $\begin{align*} \lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3} &= \lim_{x \to 3} \frac{(x - 3)(x + 1)}{x - 3} \\ & = \lim_{x \to 3} \frac{\cancel{(x - 3)}(x+1)}{\cancel{x-3}} \\ & = \lim_{x \to 3} x + 1 \\ & = (3) + 1 \\ & = 4 \end{align*}$ $\lim_{x \to 2} \frac{x^2 - 4}{x -2} = \lim_{x \to 2} \frac{(x + 2)(x-2)}{x-2} = \lim_{x \to 2} x + 2 = \boxed{4}$ $\lim_{x \to -1} \frac{x^2+6x + 5}{x +1} = \lim_{x \to -1} \frac{(x + 1)(x + 5)}{x+1} = \lim_{x \to -1} x + 5 = \boxed{4}$

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