Opérations sur les limites

α %alpha désigne un réel fini, + ∞ + infinity ou − ∞ - infinity .

Fondamental

Soient  f f et  g g deux fonctions qui admettent respectivement  l l et l ' l' (finies) lorsque  x x tend vers α %alpha , alors

  1. lim x → α f ( x ) + g ( x ) = l + l ' lim from {x toward %alpha} f(x)+g(x) = l+l'

  2. lim x → α kf ( x ) = kl lim from {x toward %alpha} kf(x) = kl quelque soit k k réel

  3. lim x → α f ( x ) g ( x ) = l l ' lim from {x toward %alpha} f(x) g(x) = l l'

  4. lim x → α f ( x ) g ( x ) = l l ' lim from {x toward %alpha} {{f(x)} over {g(x)} } = l over {l'}  si l ' ≠ 0 l'<>0

Fondamental

Si lim x → α f ( x ) = + ∞ lim from {x toward %alpha} f(x) = +infinity et lim x → α g ( x ) = + ∞ lim from {x toward %alpha} g(x) = +infinity alors :

  1. lim x → α f ( x ) + g ( x ) = + ∞ lim from {x toward %alpha} f(x)+g(x) = +infinity

  2. lim x → α kf ( x ) = + ∞ lim from {x toward %alpha} kf(x) = +infinity si k > 0 k>0 et lim x → α kf ( x ) = − ∞ lim from {x toward %alpha} kf(x) = -infinity si k < 0 k<0

  3. lim x → α f ( x ) g ( x ) = + ∞ lim from {x toward %alpha} f(x)g(x) = +infinity

Fondamental

Si lim x → α f ( x ) = l lim from {x toward %alpha} f(x) = l ( l l réel fini) et lim x → α g ( x ) = + ∞ lim from {x toward %alpha} g(x) = +infinity alors :

  1. lim x → α f ( x ) + g ( x ) = + ∞ lim from {x toward %alpha} f(x)+g(x) = +infinity

  2. lim x → α f ( x ) g ( x ) = + ∞ lim from {x toward %alpha} f(x)g(x) = +infinity si l > 0 l>0 et lim x → α f ( x ) g ( x ) = − ∞ lim from {x toward %alpha} f(x)g(x) = -infinity si l < 0 l<0

  3. lim x → α f ( x ) g ( x ) = 0 lim from {x toward %alpha} {{f(x)} over {g(x)} } = 0 si l ≠ 0 l<>0

Fondamental

l l  est un réel fini,

  1. Si lim x → α f ( x ) = l lim from {x toward %alpha} f(x) = l et lim x → α g ( x ) = 0 + lim from {x toward %alpha} g(x) = 0^{{}+{}} alors lim x → α f ( x ) g ( x ) = lim from {x toward %alpha} {{f(x)} over {g(x)} } = {} (signe de l l ) ∞ infinity

  2. Si lim x → α f ( x ) = l lim from {x toward %alpha} f(x) = l et lim x → α g ( x ) = 0 − lim from {x toward %alpha} g(x) = 0^{{}-{}} alors lim x → α f ( x ) g ( x ) = lim from {x toward %alpha} {{f(x)} over {g(x)} } = {} (signe de − l -l ) ∞ infinity

Remarque

Il existe plusieurs cas, appelés formes indéterminées :

  1. Si f ( x ) → + ∞ f(x) toward {}+{}infinity et g ( x ) → − ∞ g(x) toward {}-{}infinity , on ne peut pas conclure immédiatement pour la limite de f ( x ) + g ( x ) f(x)+g(x)

  2. Si f ( x ) → + ∞ f(x) toward {}+{}infinity (ou − ∞ - infinity ) et g ( x ) → 0 g(x) toward 0 , on ne peut pas conclure immédiatement pour la la limite de f ( x ) g ( x ) f(x)g(x)

  3. Si f ( x ) → + ∞ f(x) toward {}+{}infinity (ou − ∞ - infinity ) et g ( x ) → + ∞ g(x) toward {}+{}infinity (ou − ∞ - infinity ), on ne peut pas conclure immédiatement pour la limite de f ( x ) g ( x ) {f(x)} over {g(x)}

  4. Si f ( x ) → 0 f(x) toward 0 et g ( x ) → 0 g(x) toward 0 , on ne peut pas conclure immédiatement pour pour la limite de f ( x ) g ( x ) {f(x)} over {g(x)}

Exemple

Soit f ( x ) = ln x f(x)= ln x et g ( x ) = x g(x)= x .

Déterminer lim x → + ∞ f ( x ) − g ( x ) lim from {x toward +infinity} f(x)-g(x)

On ne peut conclure directement, on est en présence d'une forme indéterminée ∞ infinity − ∞ - infinity .

On écrit : f ( x ) − g ( x ) = ln x − x = x ( ln x x − 1 ) f(x)-g(x)= ln x - x = x({ln x} over x -1)

Donc lim x → + ∞ f ( x ) − g ( x ) = lim x → + ∞ ( − x ) lim from {x toward +infinity} f(x)-g(x)=lim from {x toward +infinity} (-x) , car lim x → + ∞ ln x x = 0 lim from {x toward +infinity} {{ln x} over x} = 0

Et donc lim x → + ∞ ln x − x = − ∞ lim from {x toward +infinity} {ln x - x} = -infinity