Craig Guerrero
09/30/2023 · Primary School
1) Sout \( x>0 \). En appliquant le thécorème des accroissement fines à la fonction \( \ln \) montrer que: \( \exists c \in] x, x+1\left[\frac{1}{c}=\ln \left(1+\frac{1}{x}\right)\right. \) 2) \( M q: \exists c \in] 0, x\left[\right. \) arctan \( (x)=\frac{x}{1+c^{2}} \) 3) En déduire que \( \forall x>0: \) \( \frac{1}{1+x}<\ln \left(1+\frac{1}{x}\right)<\frac{1}{x} \)
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For \( x > 0 \), there exists a \( c \) in \( ]x, x+1[ \) such that \( \frac{1}{c} = \ln\left(1 + \frac{1}{x}\right) \). Additionally, it holds that \( \frac{1}{1 + x} < \ln\left(1 + \frac{1}{x}\right) < \frac{1}{x} \) for all \( x > 0 \).
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