Skip to content. | Skip to navigation

Personal tools
Log in Register
Sections
You are here: Home Test Department Heildunarreglur Regla l'Hopital l'Hopital 2

l'Hopital 2

Slide View     


Main text Main image Examples
$ f(a) = g(a) = 0$ ;     $ f(x)$ og $ g(x)$ eru diffranleg á opnu bili $ I$ þ.a. $ a\in I$ . Jafnframt er $ g'(x) \neq 0$ á $ I$ ef $ x\neq a$ .
Þá gildir:

$\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}
$

(sjá Appendix 6 í TC).


Dæmi:
  1. $\displaystyle \lim_{x\to 0} \frac{1 - \cos x}{x^2} = \lim_{x\to 0} \frac{\sin x}{2x} = \lim_{x\to 0} \frac{\cos x}{2} = \frac{1}{2}
$

    (ath. l'Hopital gefur

    $\displaystyle \lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\cos x}{1} = 1
$

    eins og við höfum áður séð).
  2. (sjá example 2, á bls. 579 í TC).
        $\displaystyle \lim_{x\to 0} \frac{\sqrt{1+x} - 1 - \frac{x}{2}}{x^2} \qquad \qquad \frac{0}{0}$  
    $\displaystyle =$   $\displaystyle \lim_{x\to 0} \frac{\frac{1}{2}(1+x)^{-\frac{1}{2}} - \frac{1}{2}}{2x}$   aftur$\displaystyle \quad \frac{0}{0}$  
    $\displaystyle =$   $\displaystyle \lim_{x\to 0} \frac{\left( -\frac{1}{2} \right) \cdot \frac{1}{2} (1+x)^{-\frac{3}{2}}}{2} = -\frac{1}{8}$  



Explanation text Explanation image Details
$ f(a) = g(a) = 0$ ;     $ f(x)$ og $ g(x)$ eru diffranleg á opnu bili $ I$ þ.a. $ a\in I$ . Jafnframt er $ g'(x) \neq 0$ á $ I$ ef $ x\neq a$ .
Þá gildir:

$\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}
$

(sjá Appendix 6 í TC).

Hver er munurinn á l'Hopital 1 og 2?


References Handout Alternative

Grades may be recorded and used anonymously for research purposes.