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Suppose \(f\) is defined and differentiable for every \(x>0\), and \(f^{\prime}(x) \rightarrow 0\) as \(x \rightarrow+\infty\). Put \(g(x)=f(x+1)-f(x)\). Prove that \(g(x) \rightarrow 0\) as \(x \rightarrow+\infty\).
in Mathematics by Diamond (66,769 points) | 72 views

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Let \(\varepsilon>0\). Choose \(x_{0}\) such that \(\left|f^{\prime}(x)\right|<\varepsilon\) if \(x>x_{0}\). Then for any \(x \geq x_{0}\) there exists \(x_{1} \in(x, x+1)\) such that
f(x+1)-f(x)=f^{\prime}\left(x_{1}\right) .
Since \(\left|f^{\prime}\left(x_{1}\right)\right|<\varepsilon\), it follows that \(|f(x+1)-f(x)|<\varepsilon\), as required.
by Diamond (66,769 points)

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