Einsidige grenseverdiar

Fire nye døme

Hans Georg Schaathun

Mai 2016

$$ \begin{align} f(x) &= \begin{cases} 1, \quad\text{der } x>0,\\ 0, \quad\text{der } x=0,\\ -1, \quad\text{der } x<0. \end{cases} \end{align} $$

$$\lim_{x\to0^+} f(x) = +1$$

$$\lim_{x\to0^-} f(x) = -1$$

$\displaystyle\lim_{x\to0} f(x)$ er udefinert

$$ \begin{align} &f(x) = \sqrt x\\ &\lim_{x\to0} f(x) = ?? \end{align} $$

$$\lim_{x\to0^+} f(x) = \sqrt0 = 0$$

$\displaystyle\lim_{x\to0^-} f(x)$ er udefinert

$\displaystyle\lim_{x\to0} f(x)$ er udefinert

$$f(x) = \frac{1}{x}$$

$$\lim_{x\to0^+} f(x) = \infty$$

$$\lim_{x\to0^-} f(x) = -\infty$$

$$\lim_{x\to0} f(x) = \pm\infty$$

$$ \begin{align} f(x) &= \frac{x^2-1}{x-1} = \frac{(x+1)(x-1)}{x-1} \\ f(x) &= x+1,\quad x\neq1 \end{align} $$

$$ \begin{align} \lim_{x\to1^+} f(x) = (x+1)\big|_{x=1} = 2,\\ \lim_{x\to1^-} f(x) = (x+1)\big|_{x=1} = 2,\\ \lim_{x\to1} f(x) = (x+1)\big|_{x=1} = 2 \end{align} $$