Limit of series of functions Let $x\in[0,\infty)$ and $$f_n(x)=\frac{nx}{nx+1}$$ Find the pointwise limit for $x\in[0,\infty)$, and show that it is only uniform on $[1,\infty)$ and not on $[0,\infty)$. 
Intuitively the limit would be $1$ as $n\to\infty$, and I can show that when $n\geq \frac{\frac{1}{\varepsilon}-1}{x}$, $|f_n(x)-f(x)|<\varepsilon$. But I'm not sure how to prove the uniformity part.
 A: Let $\epsilon \gt 0$ be given, and let $x\ge 1$.
Note that $\frac{nx}{nx+1}=1-\frac{1}{nx+1}$. This is within $\epsilon$ of $1$ whenever $\frac{1}{nx+1}\lt \epsilon$, and since $x\ge 1$, this holds whenever $\frac{1}{n+1}\lt \epsilon$. So we will be within $\epsilon$ of whenever $n$ is a positive integer $\gt \frac{1}{\epsilon}$. The bound is independent of $x$.
A minor modification proves uniform convergence in the interval $[a,\infty)$, where $a$ is any fixed positive real.
We do not have uniform convergence in $[0,\infty)$. We leave this to you, at least for a while. The shortest proof uses a property of the limit of a uniformly convergent sequence of continuous functions.
A: It helps to realize that the choice of $[1,\infty)$ is somewhat contrived. Let $\varepsilon > 0$ be given, and note that $x \geq 1$ implies $1/x \leq 1$, so that
$$\left|\frac{nx}{nx+1}-1  \right| = \left|\frac{-1}{nx+1}  \right| \leq \frac1n \left| \frac1x \right| \leq \varepsilon.$$
Showing that convergence is not uniform on $[0,\infty)$ is a matter of realizing $1/x$ does not behave well near 0. Indeed, when $x = 0$, the $f_n(x) = 0$ for all $n$, so that the pointwise limit is 0 when $x = 0$ and 1 when $x > 0$. If we let $f$ be the pointwise limit, then 
$$\sup_{x \geq 0} |f_n(x) - f(x)| = 1.$$
