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 $\lim_{x \rightarrow +\infty} f(x)$ .

#Calculus #Limits #Finn #CanYouProveIt? #DifferentiableFunction

Easy Math Editor

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Comments

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 - 7 years, 1 month ago

Awesome. :D

Finn Hulse - 7 years, 1 month ago

Nice proof.

Sharky Kesa - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$