Calcula los límites
| $\begin{array}{ll}\text{1)} &\displaystyle \lim_{x \to \infty} e^{5x^3+4x^2+5x-1} \\ \\ \text{2)} &\displaystyle \lim_{x \to -\infty}\displaystyle \frac {x^4+4x^2-5}{-x^2-3x-6} \\\\ \text{3)} &\displaystyle \lim_{x \to \infty} \displaystyle \frac {-8x+4}{7x^2-8x} \\\\ \text{4)} &\displaystyle \lim_{x \to -\infty}\displaystyle \frac {6x^2+x}{10x^2-4} \\\\ \text{5)} &\displaystyle \lim_{x \to -1}\displaystyle \frac {9x^2-x^3}{x^2-1} \end{array}$ |
$\begin{array}{ll}\text{6)} &\displaystyle \lim_{x \to -5}\displaystyle \frac {-1}{(x+5)^2}\\ \\ \text{7)} &\displaystyle \lim_{x \to -\infty}\displaystyle \left ( \frac{3}{4} \right )^{-x} \\\\ \text{8)} &\displaystyle \lim_{x \to \infty} \displaystyle \ln \left (\frac {-8x^3+4}{7x^2-8x} \right ) \\\\ \text{9)} &\displaystyle \lim_{x \to 0}\displaystyle \frac {x}{x+4} \\\\ \text{10)} &\displaystyle \lim_{x \to 1}\displaystyle \frac {x+1}{3x-3} \end{array}$ |
Soluciones
Pulsa en las pestañas para ver las soluciones.
1
$\displaystyle \lim_{x \to \infty} f(x) = \infty$
2
$\displaystyle \lim_{x \to -\infty} f(x) = -\infty$
3
$\displaystyle \lim_{x \to \infty} f(x) = 0$
4
$\displaystyle \lim_{x \to -\infty} f(x) = \displaystyle\frac {6}{10} = \displaystyle\frac {3}{5}$
Es mejor dejarlo en fracción pero también se admitirá 0.6
5
$\displaystyle \lim_{x \to -1^-} f(x) = +\infty$
$\displaystyle \lim_{x \to -1^+} f(x) = -\infty$
$\displaystyle \lim_{x \to -1} f(x) = \pm \infty$
6
$\displaystyle \lim_{x \to -5^-} f(x) = -\infty$
$\displaystyle \lim_{x \to -5^+} f(x) = -\infty$
$\displaystyle \lim_{x \to -5} f(x) = -\infty$
7
$\displaystyle \lim_{x \to -\infty} f(x) = 0$
8
$\displaystyle \lim_{x \to \infty} f(x) = \emptyset$
9
$\displaystyle \lim_{x \to 0} f(x) = 0$
10
$\displaystyle \lim_{x \to 1^-} f(x) = -\infty$
$\displaystyle \lim_{x \to 1^+} f(x) = +\infty$
$\displaystyle \lim_{x \to 1} f(x) = \pm \infty$