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Miscel·lània 1

Calcula els límits

$\begin{array}{ll}\text{1)} &\displaystyle \lim_{x \to \infty} e^{5x^3+4x^2+5x-1} \\ \\
\text{2)} &\displaystyle \lim_{x \to -\infty}\displaystyle \frac {x^4+4x^2-5}{-x^2-3x-6} \\\\
\text{3)} &\displaystyle \lim_{x \to \infty} \displaystyle \frac {-8x+4}{7x^2-8x} \\\\
\text{4)} &\displaystyle \lim_{x \to -\infty}\displaystyle \frac {6x^2+x}{10x^2-4} \\\\
\text{5)} &\displaystyle \lim_{x \to -1}\displaystyle \frac {9x^2-x^3}{x^2-1} \end{array}$
$\begin{array}{ll}\text{6)} &\displaystyle \lim_{x \to -5}\displaystyle \frac {-1}{(x+5)^2}\\ \\
\text{7)} &\displaystyle \lim_{x \to -\infty}\displaystyle \left ( \frac{3}{4}  \right )^{-x} \\\\
\text{8)} &\displaystyle \lim_{x \to \infty} \displaystyle \ln \left (\frac {-8x^3+4}{7x^2-8x} \right ) \\\\
\text{9)} &\displaystyle \lim_{x \to 0}\displaystyle \frac {x}{x+4} \\\\
\text{10)} &\displaystyle \lim_{x \to 1}\displaystyle \frac {x+1}{3x-3} \end{array}$


Solucions

Fes clic a les pestanyes per veure les solucions.

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

$\displaystyle \lim_{x \to -\infty} f(x) = \displaystyle\frac {6}{10} = \displaystyle\frac {3}{5}$

És millor deixar-ho en fracció però també s'admetrà 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

$\displaystyle \lim_{x \to 1^-} f(x) = -\infty$

$\displaystyle \lim_{x \to 1^+} f(x) = +\infty$

$\displaystyle \lim_{x \to 1} f(x) = \pm \infty$