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Proving that a limit exists

Let \(f: [0, +\infty) \rightarrow \mathbb{R}\) be a differentiable function. Suppose that \(|f(x)| \leq 5\) and \(f(x)f'(x)>\sin{x}\) for all \(x \geq 0\). Prove that there exists \(\lim_{x \rightarrow +\infty} f(x)\).

Note by Finn Hulse
2 years, 8 months ago

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Counter-example : If such a function \(f(x)\) indeed exists, consider the problem in terms of the function \(g(x)=\frac{1}{2}f^2(x)\). Then the problem asks us to find a differentiable function function \(g:[0, \infty)\to \mathbb{R}\) such that \(0\leq g(x) \leq 25/2\) and \(g'(x) > \sin x\) with the property that \(\lim_{x\to +\infty} g(x) \) exists. However, the following function \(g(x)=10-\exp(-x)-\cos(x) \) clearly satisfies the first two conditions and violates the last limiting condition as \(\lim_{x\to +\infty} g(x)\) does not exist.

Hence \(f(x)=\sqrt{2(10-\exp(-x)-\cos(x))}\) constitutes a counter-example of the original problem. Abhishek Sinha · 2 years, 8 months ago

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@Abhishek Sinha Nice proof. Sharky Kesa · 2 years, 8 months ago

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@Abhishek Sinha Awesome. :D Finn Hulse · 2 years, 8 months ago

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