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youyou19987
@youyou19987
January 2021
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Bonsoir , la solution sans utiliser la règle d'hôpital et merci
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scoladan
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Bonjour,
xln(1+e^x) = ln(1+e^x)^x
quand x --> - infini,
e^x --> 0+
==> 1+e^x --> 1+
==> (1+e^x)^x --> 1^x = 1-
==> ln(1+e^x)^x --> ln(1-) = 0-
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Lista de comentários
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Bonjour,xln(1+e^x) = ln(1+e^x)^x
quand x --> - infini,
e^x --> 0+
==> 1+e^x --> 1+
==> (1+e^x)^x --> 1^x = 1-
==> ln(1+e^x)^x --> ln(1-) = 0-