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l'Hopital 1

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$ f(a) = g(a) = 0$ ;     $ f'(a)$ og $ g'(a)$ eru til og $ g'(a) \neq 0$ . Þá er

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






$\displaystyle \lim_{x\to 1} \frac{x^3 -1}{4x^3 - x - 3} = \frac{3x^2\big\vert _{x=1}}{12x^2 - 1\big\vert _{x=1}} = \frac{3}{12-1} = \frac{3}{11}
$



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