Re: Sequences of Functions

Suppose you know that $\displaystyle \arctan(x)\to\pi/2$ as $\displaystyle x\to\infty$. That is, there is a function $\displaystyle \mu$ mapping neighborhoods of$\displaystyle \pi/2$ to neighborhoods of infinity: for every $\displaystyle \varepsilon>0$ and every $\displaystyle x>\mu(\varepsilon)$ we have $\displaystyle |\arctan(x)-\pi/2|<\varepsilon$.

Now fix an arbitrary x > 0. Let $\displaystyle N(\varepsilon)=\lceil\mu(\varepsilon)/x\rceil$ where $\displaystyle \lceil\cdot\rceil$ is the ceiling function. The essential property of $\displaystyle N(\varepsilon)$ is that $\displaystyle n>N(\varepsilon)$ implies $\displaystyle nx>\mu(\varepsilon)$. Therefore, for any $\displaystyle \varepsilon>0$ and $\displaystyle n>N(\varepsilon)$ we have $\displaystyle |\arctan(nx)-\pi/2|<\varepsilon$.

This (and a similar argument for x < 0) proves that $\displaystyle \arctan(nx)$ converges to $\displaystyle (\pi/2)\mathop{\text{sgn}}(x)$ pointwise. Note that it does not converge uniformly because $\displaystyle \sup_{x\in\mathbb{R}} |\arctan(nx)-\pi/2|= 1$ for all n.