TITLE: uniform convergence of few sequence of functions
QUESTION [7 upvotes]: Pick out the sequences $\{f_n\}$ which are uniformly convergent.
(a) $f_n(x) = nxe^{−nx}$ on $(0,∞)$.
(b)$f_n(x) = x^n$ on $[0, 1]$.
(c)$f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$ on $\mathbb{R}$.
(d) $f_n(x)=\frac{nx}{1 + nx}$ on $(0,1)$
(e) $f_n(x) = ∑_{n=1}^∞\frac{n\sin(nx)}{e^n}$ on $[0,\pi]$ 
(f) $f_n(x) = \frac{x^n}{1 + x^n}$ ; on  $[0, 2]$ 
(g) $f_n(x) = \sin^nx$ on  $[0,\pi/2)$    
(h) $f_n(x) = (x^n/n)+1$ on $[0,1)$
(i) $f_n(x) = \frac{1}{1+(x-n)^2}$ on $(-∞,0)$  
(j) $f_n(x) = \frac{1}{1+(x-n)^2}$ on $(0, ∞)$    
(a) Here the function becomes $nx/e^{nx}$ which is tends to $0$ as $n$ tends to $∞$ so the function is uniformly continuous. I am not sure though.
(b) Here limit function is not continuous so not uniformly continuous
(c) true by same reason of (a)
(d)  false as $n$ tends to $∞$ the function tends to $1$ not zero.
(e) no idea
(f) false as limit function is not continuous
(g) true by M test
(h) true by M test
(i) no idea
(j) no idea
Can somebody guide me properly please

REPLY [4 votes]: A related technique. Note that, for (e), you can use the Weierstrass M-Test, since
$$\left| \frac{n\sin(nx)}{e^n} \right| \leq  \left| \frac{n}{e^n} \right|=\frac{n}{e^n}, $$
and
$$ \sum_{n=1}^{\infty}\frac{n}{e^n} < \infty.  $$
Note: if 
$$\lim_{ n \to \infty}f_n(x)=f(x) \quad\mathrm{ and } \quad \sup| f_n(x)-f(x) | < \epsilon, $$
then the sequence converges uniformly.
For (j), notice that, $f_n(x)\to 0$ for $x\neq n$ and $f_n(x)\to 1$ for $x = n$ which implies
$$ \sup |f_n(x)-f(x)| =1 \nless \epsilon. $$