# Rasjonale funksjonar mot uendeleg

## Fleire grenseverdiar med løysing

Juni 2016

$f\left(x\right)=\frac{{x}^{3}-1000{x}^{2}-100x}{-\frac{{x}^{4}}{1000}+{x}^{3}-1000{x}^{2}-100x}$
$\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{3}-1000{x}^{2}-100x}{-\frac{{x}^{4}}{1000}+{x}^{3}-1000{x}^{2}-100x}$
$f\left(x\right)=\frac{{x}^{3}-1000{x}^{2}-100x}{-\frac{{x}^{4}}{1000}+{x}^{3}-1000{x}^{2}-100x}$

$$\lim_{x\to\infty} f(x) = 0$$

$$\lim_{x\to-\infty} f(x) = 0$$

$$g(x) = \frac{-\frac{x^4}{1000} + x^3 - 1000x^2 - 100x} {x^3 - 1000x^2 - 100x}$$
$$x\to\pm\infty$$

$$g(x) = \frac{-\frac{x^4}{1000} + x^3 - 1000x^2 - 100x} {x^3 - 1000x^2 - 100x}$$

$$g(x) = \frac {-\frac{x^4}{1000x^4} + \frac{x^3}{x^4} - 1000\frac{x^2}{x^4} - \frac{100x}{x^4}} {\frac{x^3}{x^4} - \frac{1000x^2}{x^4} - \frac{100x}{x^4}}$$

$$g(x) = \frac{-\frac{1}{1000} + \frac{1}{x} - 1000\frac{1}{x^2} - \frac{100}{x^3}} {\frac{1}{x} - \frac{1000}{x^2} - \frac{100}{x^3}}$$

$$g(x)\to\pm\infty$$

$$g(x) = \frac{-\frac{x^4}{1000} + x^3 - 1000x^2 - 100x} {x^3 - 1000x^2 - 100x}$$
\begin{align} \lim_{x\to\infty} g(x) = -\infty \\ \lim_{x\to-\infty} g(x) = \infty \end{align}
$$h(x) = \frac{x^3 - 10x^2 - 2x} {x^3 + 5x^2 - 2}$$
$$x\to\pm\infty$$

$$h(x) = \frac{x^3 - 10x^2 - 2x}{x^3 + 5x^2 - 2}$$

$$h(x) = \frac{x^3/x^3 - 10x^2/x^3 - 2x/x^3}{x^3/x^3 + 5x^2/x^3 - 2/x^3}$$

$$h(x) = \frac{1 - 10/x - 2/x^2}{1 + 5/x - 2/x^3}$$

$$h(x) = \frac{x^3 - 10x^2 - 2x} {x^3 + 5x^2 - 2}$$
\begin{align*} \lim_{x\to\infty} h(x) = 1 \\ \lim_{x\to{-\infty}} h(x) = 1 \end{align*}

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

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