# 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
4 years, 1 month 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.

- 4 years, 1 month ago

Nice proof.

- 4 years, 1 month ago

Awesome. :D

- 4 years, 1 month ago