Fonction logarithme neperien - Examen corrigé
Fonction logarithme neperien - Examen corrigé
Fonction logarithme neperien - Examen corrigé
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<strong>corrigé</strong> exercice 11 :<br />
déterminer les limites suivantes :<br />
a. lim<br />
x→+∞ (x2 +lnx)<br />
lim<br />
x→+∞ (x2 +lnx) = +∞+(+∞) = +∞ (par addition)<br />
b. lim<br />
x→+∞ (1−lnx)lnx<br />
lim (1−lnx)lnx = (1−(+∞))×(+∞) = (−∞)×(+∞) = −∞ (par multiplication)<br />
x→+∞<br />
c. lim<br />
x→+∞ (ln2−3lnx)<br />
lim (ln2−3lnx) = ln2−3×(+∞) = ln2−∞ = −∞ (par addition)<br />
x→+∞<br />
d. lim<br />
x→+∞ (1−x2 )(lnx+1)<br />
lim<br />
x→+∞ (1−x2 )(lnx+1) = (1−∞)×(+∞+1) = (−∞)×(+∞) = +∞ (par multiplication)<br />
<strong>corrigé</strong> exercice 12 :<br />
déterminer les limites suivantes :<br />
a. lim<br />
x→o +(lnx+x+1)<br />
lim = −∞+0+1 = −∞ (par addition)<br />
x→o +(lnx+x+1)<br />
b. lim<br />
x→o +(1 x −4lnx)<br />
lim<br />
x→o +(1<br />
−4lnx) = +∞−4×(−∞) = +∞+(+∞) = +∞ (par addition)<br />
x<br />
c. lim<br />
x→o +<br />
2lnx+1<br />
x<br />
lim<br />
x→o +<br />
2lnx+1<br />
=<br />
x<br />
2×(−∞)+1<br />
0 +<br />
d. lim<br />
x→o +(−x+1−lnx)<br />
= −∞<br />
= −∞ (par quotient)<br />
0 +<br />
lim = 0+1−(−∞) = +∞ (par addition)<br />
x→o +(−x+1−lnx)<br />
<strong>corrigé</strong> exercice 13 :<br />
déterminer les limites suivantes :<br />
lnx<br />
b. lim (4+<br />
x→+∞ x )<br />
lnx<br />
lim (4+ ) = 4+0 = 4 (par addition)<br />
x→+∞ x