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A4-format til udskrift. - Aarhus Universitet

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112 III. POTENSRÆKKER<br />

kaldes ubestemt af form ∞<br />

∞ .<br />

Eksempel<br />

er ubestemt af form ∞<br />

∞ .<br />

ln |x|<br />

lim<br />

x→0 x−1 1.5. l’Hospitals regel 1 ☞ [S] 4.5 Indeterminate forms and l’Hospital’s rule<br />

Sætning (l’Hospitals regel)<br />

Antag, at f,g er differentiable og g ′ (x) = 0 for x = a <strong>til</strong>pas nœr a. Hvis<br />

er et ubestemt udtryk af form 0<br />

0 , så er<br />

f(x)<br />

lim<br />

x→a g(x)<br />

f(x) f<br />

lim = lim<br />

x→a g(x) x→a<br />

′ (x)<br />

g ′ (x)<br />

1.6. Overbevis ☞ [S] 4.5 Indeterminate forms and l’Hospital’s rule<br />

Bevis<br />

Fra den udvidede middelværdisætning<br />

Beregn nu<br />

f(x)g ′ (x ∗ ) = f ′ (x ∗ )g(x), a < x ∗ < x<br />

f(x) f<br />

lim = lim<br />

x→a g(x) x→a<br />

′ (x∗ )<br />

g ′ (x∗ f<br />

= lim<br />

) x→a<br />

′ (x)<br />

g ′ (x)<br />

1.7. Prøv reglen ☞ [S] 4.5 Indeterminate forms and l’Hospital’s rule<br />

Eksempel<br />

ubestemt af form 0<br />

0 .<br />

Heraf fås<br />

x<br />

lim<br />

x→1<br />

2 − 1<br />

x − 1<br />

f(x) = x 2 − 1,f ′ (x) = 2x<br />

g(x) = x − 1,g ′ (x) = 1<br />

x<br />

lim<br />

x→1<br />

2 − 1 2x<br />

= lim = 2<br />

x − 1 x→1 1<br />

1.8. l’Hospitals regel 2 ☞ [S] 4.5 Indeterminate forms and l’Hospital’s rule<br />

Sætning (l’Hospitals regel)<br />

Antag, at f,g er differentiable og g ′ (x) = 0 for x = a <strong>til</strong>pas nœr a. Hvis<br />

er et ubestemt udtryk af form ∞<br />

∞ , så er<br />

f(x)<br />

lim<br />

x→a g(x)<br />

f(x) f<br />

lim = lim<br />

x→a g(x) x→a<br />

′ (x)<br />

g ′ (x)

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