Higher education
Preparatory classes, engineering schools, medicine, business: the subjects of your year, semester by semester.
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Higher education · High school · Programming · Data Science & AI
Understand turns difficult courses into ideas you can explain. Clear lessons, exercises that build step by step, hints when you are stuck and corrections that show the reasoning — not just the answer.
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On veut calculer :
\[L = \lim_{x \to +\infty} x\left(\frac{\pi}{2} - \arctan x\right)\]
C'est une forme « \(\infty \times 0\) ».
QCalculer \(\displaystyle\lim_{t \to 0} \frac{\arctan t}{t}\).
0not quite
Numérateur et dénominateur tendent vers 0 : forme indéterminée.
Pose \(u = \arctan t\). Alors \(t = \tan u\), et quand \(t \to 0\), \(u \to \dots\)
\(\dfrac{\arctan t}{t} = \dfrac{u}{\tan u}\).
Posons \(u = \arctan t\), donc \(t = \tan u\). Pour \(t \neq 0\) : \(\dfrac{\arctan t}{t} = \dfrac{u}{\tan u}\).
\(\arctan\) est continue en \(0\) et \(\arctan 0 = 0\) : quand \(t \to 0\), \(u \to 0\) (avec \(u \neq 0\)).
Or \(\displaystyle\lim_{u \to 0} \frac{\tan u}{u} = 1\), donc \(\displaystyle\lim_{u \to 0} \frac{u}{\tan u} = 1\).
Par composition : \(\displaystyle\lim_{t \to 0} \frac{\arctan t}{t} = 1\).
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Preparatory classes, engineering schools, medicine, business: the subjects of your year, semester by semester.
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