↑ stuart clark:
Hi.
Skrytý text:
Since [mathjax]g[/mathjax] is bounded, we have [mathjax]g(x)\leq\|g\|_{\infty}[/mathjax] for all [mathjax]x\in\mathbb{R}[/mathjax].
Consequently
[mathjax] G(x) = G(0)+\int_0^xg(t)\,\mathrm{d}t = \int_0^xg(t)\,\mathrm{d}t = \int_0^{3\left \lfloor \frac{x}{3} \right \rfloor} g(t)\,\mathrm{d}t + \int_{3\left \lfloor \frac{x}{3} \right \rfloor}^x g(t)\,\mathrm{d}t = 6\left \lfloor \frac{x}{3} \right \rfloor + \int_{3\left \lfloor \frac{x}{3} \right \rfloor}^x g(t)\,\mathrm{d}t[/mathjax]
Since [mathjax]\frac{x}{3}-1\leq \left \lfloor \frac{x}{3} \right \rfloor\leq \frac{x}{3}[/mathjax], we obtain the estimates
[mathjax]2x-6 - 3\|g\|_{\infty} \leq G(x) \leq 2x + 3\|g\|_{\infty}[/mathjax].
Hence [mathjax]|G(x)-2x|\leq 6+3\|g\|_{\infty}[/mathjax] for all [mathjax]x\in\mathbb{R}[/mathjax] and therefore
[mathjax]\left|\lim_{x\to0}xG\left(\frac{1}{x}\right) - 2\right| = \left|\lim_{x\to0}x\left(G\left(\frac{1}{x}\right)-\frac{2}{x}\right)\right| = \lim_{x\to0}|x|\left|G\left(\frac{1}{x}\right)-\frac{2}{x}\right| \leq \lim_{x\to0}|x|( 6+3\|g\|_{\infty}) = 0 [/mathjax]
Thus [mathjax]\lim_{x\to0}xG\left(\frac{1}{x}\right)=2[/mathjax].