Problema 12

Calcular los siguientes límites si es que existen:

1. $lim_{x \to 3} \frac{x^{2} - 9}{x - 3}$

\begin{aligned} lim_{x \to 3} \frac{x^{2} - 9}{x - 3} & = lim_{x \to 3} \frac{(x + 3) (x - 3)}{x - 3} \\ = lim_{x \to 3} x + 3 \\ = 3 + 3 = 6 \end{aligned}

lim_{x \to - 1} \frac{x + 1}{x^{3} + 1}

\begin{aligned} lim_{x \to - 1} \frac{x + 1}{x^{3} + 1} & = lim_{x \to - 1} \frac{x + 1}{(x + 1) (x^{2} - x + 1)} \\ = lim_{x \to - 1} \frac{1}{x^{2} - x + 1} \\ = \frac{1}{(- 1)^{2} - (- 1) + 1} = \frac{1}{3} \end{aligned}

lim_{x \to 2} \frac{x - 2}{x^{2} - 4}

\begin{aligned} lim_{x \to 2} \frac{x - 2}{x^{2} - 4} & = lim_{x \to 2} \frac{x - 2}{(x + 2) (x - 2)} \\ = lim_{x \to 2} \frac{1}{x + 2} \\ = \frac{1}{2 + 2} = \frac{1}{4} \end{aligned}

lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h}

\begin{aligned} lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h} & = lim_{h \to 0} \frac{x^{2} + 2 x h + h^{2} - x^{2}}{h} \\ = lim_{h \to 0} \frac{2 x h + h^{2}}{h} \\ = lim_{h \to 0} 2 x + h \\ = 2 x + 0 = 2 x \end{aligned}

lim_{x \to - 2} \frac{3 x^{2} + 2 x - 8}{x + 2}

** Trabajo previo:

3 x^{2} + 2 x - 8 = \frac{(3 x)^{2} + 2 (3 x) - 24}{3} = \frac{(3 x + 6) (3 x - 4)}{3} = (x + 2) (3 x - 4)

\begin{aligned} lim_{x \to - 2} \frac{3 x^{2} + 2 x - 8}{x + 2} & = lim_{x \to - 2} \frac{(x + 2) (3 x - 4)}{x + 2} \\ = lim_{x \to - 2} 3 x - 4 \\ = 3 (- 2) - 4 \\ = - 10 \end{aligned}

lim_{x \to 2} \frac{\sqrt{2 x - 3} - \sqrt{x - 1}}{x^{2} - 4}

\begin{aligned} lim_{x \to 2} \frac{\sqrt{2 x - 3} - \sqrt{x - 1}}{x^{2} - 4} & = lim_{x \to 2} \frac{\sqrt{2 x - 3} - \sqrt{x - 1}}{x^{2} - 4} \cdot \frac{\sqrt{2 x - 3} + \sqrt{x - 1}}{\sqrt{2 x - 3} + \sqrt{x - 1}} \\ = lim_{x \to 2} \frac{2 x - 3 - x + 1}{(x^{2} - 4) (\sqrt{2 x - 3} + \sqrt{x - 1})} \\ = lim_{x \to 2} \frac{x - 2}{(x + 2) (x - 2) (\sqrt{2 x - 3} + \sqrt{x - 1})} \\ = lim_{x \to 2} \frac{1}{(x + 2) (\sqrt{2 x - 3} + \sqrt{x - 1})} \\ = \frac{1}{(2 + 2) (\sqrt{2 (2) - 3} + \sqrt{2 - 1})} \\ = \frac{1}{(4) (\sqrt{1} + \sqrt{1})} = \frac{1}{8} \end{aligned}

lim_{x \to + \infty} \frac{4 x^{3} + x^{2} - 1}{5 x^{3} + x}

\begin{aligned} lim_{x \to + \infty} \frac{4 x^{3} + x^{2} - 1}{5 x^{3} + x} & = lim_{x \to + \infty} \frac{\frac{4 x^{3}}{x^{3}} + \frac{x^{2}}{x^{3}} - \frac{1}{x^{3}}}{\frac{5 x^{3}}{x^{3}} + \frac{x}{x^{3}}} \\ = lim_{x \to + \infty} \frac{4 + \frac{1}{x} - \frac{1}{x^{3}}}{5 + \frac{1}{x^{2}}} \\ = \frac{4 + \frac{1}{\infty} - \frac{1}{\infty^{3}}}{5 + \frac{1}{\infty^{2}}} = \frac{4}{5} \end{aligned}

lim_{x \to + \infty} \frac{(x + 18) (x^{4} - x)}{x^{5} + 1}

\begin{aligned} lim_{x \to + \infty} \frac{(x + 18) (x^{4} - x)}{x^{5} + 1} & = lim_{x \to + \infty} \frac{x^{5} + 18 x^{4} - x^{2} - 18 x}{x^{5} + 1} \\ = lim_{x \to + \infty} \frac{\frac{x^{5}}{x^{5}} + \frac{18 x^{4}}{x^{5}} - \frac{x^{2}}{x^{5}} - \frac{18 x}{x^{5}}}{\frac{x^{5}}{x^{5}} + \frac{1}{x^{5}}} \\ = lim_{x \to + \infty} \frac{1 + \frac{18}{x} - \frac{1}{x^{3}} - \frac{18}{x^{4}}}{1 + \frac{1}{x^{5}}} \\ = \frac{1 + \frac{18}{\infty} - \frac{1}{\infty^{3}} - \frac{18}{\infty^{4}}}{1 + \frac{1}{\infty^{5}}} = 1 \end{aligned}

lim_{x \to 1} \frac{x + 1}{x^{3} - 1}

\begin{aligned} lim_{x \to 1} \frac{x + 1}{x^{3} - 1} & = \frac{1 + 1}{1 - 1} \\ = \frac{2}{0} \end{aligned}

El límite no existe. 10.

lim_{x \to 0} \frac{1 - \cos (x)}{\sin (x)}

\begin{aligned} lim_{x \to 0} \frac{1 - \cos (x)}{\sin (x)} & = lim_{x \to 0} \frac{1 - \cos (x)}{\sin (x)} \cdot \frac{1 + \cos (x)}{1 + \cos (x)} \\ = lim_{x \to 0} \frac{1 - \cos^{2} (x)}{\sin (x) (1 + \cos (x))} \\ = lim_{x \to 0} \frac{\sin^{2} (x)}{\sin (x) (1 + \cos (x))} \\ = lim_{x \to 0} \frac{\sin (x)}{1 + \cos (x)} \\ = \frac{\sin (0)}{1 + \cos (0)} \\ = \frac{0}{1 + 1} = 0 \end{aligned}

11.

lim_{x \to 2} \frac{x^{3} - 8}{x - 2}

\begin{aligned} lim_{x \to 2} \frac{x^{3} - 8}{x - 2} & = lim_{x \to 2} \frac{(x - 2) (x^{2} + 2 x + 4)}{x - 2} \\ = lim_{x \to 2} x^{2} + 2 x + 4 \\ = 2^{2} + 2 (2) + 4 = 12 \end{aligned}

12.

lim_{x \to - \infty} \frac{3 x^{3} - 2 x^{2} + 7}{6 x^{4} - x^{3} + 2 x - 100}

\begin{aligned} lim_{x \to - \infty} \frac{3 x^{3} - 2 x^{2} + 7}{6 x^{4} - x^{3} + 2 x - 100} & = lim_{x \to - \infty} \frac{\frac{3 x^{3}}{x^{4}} - \frac{2 x^{2}}{x^{4}} + \frac{7}{x^{4}}}{\frac{6 x^{4}}{x^{4}} - \frac{x^{3}}{x^{4}} + \frac{2 x}{x^{4}} - \frac{100}{x^{4}}} \\ = lim_{x \to - \infty} \frac{\frac{3}{x} - \frac{2}{x^{2}} + \frac{7}{x^{4}}}{6 - \frac{1}{x} + \frac{2}{x^{3}} - \frac{100}{x^{4}}} \\ = \frac{\frac{3}{- \infty} - \frac{2}{(- \infty)^{2}} + \frac{7}{(- \infty)^{4}}}{6 - \frac{1}{- \infty} + \frac{2}{(- \infty)^{3}} - \frac{100}{(- \infty)^{4}}} = 0 \end{aligned}

13.

lim_{x \to \infty} \frac{x^{2} - 2 x + 5}{\sqrt{3 x^{4} + 2 x + 1}}

\begin{aligned} lim_{x \to + \infty} \frac{x^{2} - 2 x + 5}{\sqrt{3 x^{4} + 2 x + 1}} & = lim_{x \to + \infty} \frac{\frac{x^{2}}{x^{2}} - \frac{2 x}{x^{2}} + \frac{5}{x^{2}}}{\sqrt{\frac{3 x^{4}}{x^{4}} + \frac{2 x}{x^{4}} + \frac{1}{x^{4}}}} \\ = lim_{x \to + \infty} \frac{1 - \frac{2}{x} + \frac{5}{x^{2}}}{\sqrt{3 + \frac{2}{x^{3}} + \frac{1}{x^{4}}}} \\ = \frac{1 - \frac{2}{\infty} + \frac{5}{\infty^{2}}}{\sqrt{3 + \frac{2}{\infty^{3}} + \frac{1}{\infty^{4}}}} = \frac{1}{\sqrt{3}} \end{aligned}

14.

lim_{x \to y} \frac{x^{n} - y^{n}}{x - y}

\begin{aligned} lim_{x \to y} \frac{x^{n} - y^{n}}{x - y} & = lim_{x \to y} \frac{(x - y) (\sum_{k = 0}^{n - 1} x^{n - k - 1} y^{k})}{x - y} \\ = lim_{x \to y} \sum_{k = 0}^{n - 1} x^{n - k - 1} y^{k} \\ = \sum_{k = 0}^{n - 1} y^{n - k - 1} y^{k} = \sum_{k = 0}^{n - 1} y^{n - 1} = n y^{n - 1} \end{aligned}