Réponse :
Bonjour,
Explications étape par étape :
[tex]|\sqrt{x^2+1} -n| \leq \dfrac{1}{2n}\\ \\ \Longleftrightarrow\ \boxed{-\dfrac{1}{2n}\leq \sqrt{x^2+1} -n\leq \dfrac{1}{2n}}\\\\\\1.\\\\n^2+1\geq n^2\\\\\Longrightarrow\ \sqrt{n^2+1} \geq n\\\\\Longrightarrow\ \sqrt{n^2+1} -n\geq 0 \geq -\dfrac{1}{2n} \\\\\\2.\\\\0\leq \dfrac{1}{4n^2} \\\\\Longrightarrow\ n^2+1\leq n^2+1+\dfrac{1}{4n^2}\\\\\Longrightarrow\ n^2+1\leq (n+\dfrac{1}{2n})^2\\\\\Longrightarrow\ \sqrt{n^2+1} \leq n+\dfrac{1}{2n}\\\\[/tex]
[tex]\Longrightarrow\ \boxed{\sqrt{n^2+1} -n \leq \dfrac{1}{2n}}\\[/tex]
