How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$ 
How to prove uniform convergence for sequences $f_n = x^n(1-x), f_n = \frac {n^3x} {1+n^4x^2}$ and $ f_n = \sqrt {n} xe^{-nx^2}$ on $[0,1]$

I've already shown that the following sequence of functions converge pointwise to $0$ on $[0,1]$:
$$f_n = \frac {2x} {1+nx}$$
$$f_n = x^n(1-x)$$
$$f_n = \frac {n^3x} {1+n^4x^2}$$
$$f_n = \sqrt {n} xe^{-nx^2}$$
For $f_n = \frac {2x} {1+nx}$ I have $\frac {2x} {1+nx} =\frac {2} {1/x+n} \le \frac 2 n \le \epsilon$, so I can choose $N$ large enough and the inequality will hold for all $x \in [0,1]$.
However, I don't see how to prove whether or not the other three functions converge uniformly. I'm not sure I'm supposed to use majorant series, since this exercise is a basic one.
Could someone point out how to prove uniform convergence for the three sequences in question ?

I will show the 1st function $f_n(x) = x^n\cdot (1 - x)$ to converge uniformly on $[0, 1]$. $f_n'(x) = nx^{n-1} - (n+1)x^n = 0 \iff x = 0, \dfrac{n}{n+1}$. So $\displaystyle \sup_{x \in [0, 1]} |f_n(x) - 0| = f_n\left(\frac{n}{n+1}\right) = \dfrac{1}{n+1}\cdot \dfrac{1}{(1 + \frac{1}{n})^n} \to 0$ as $n \to \infty$. 
For the second function $f_n(x) = \dfrac{xn^3}{1 + n^4x^2}$, we have:
$f_n'(x) = \dfrac{n^3 - x^2n^7}{(1 + n^4x^2)^2} = 0 \iff n^3 - x^2n^7 = 0 \iff x = \dfrac{1}{n^2}$. So $\displaystyle \sup_{x \in [0,1]} |f_n(x) - 0| = f_n\left(\frac{1}{n^2}\right) = \dfrac{n}{1 + n^2} \to 0$ as $n \to \infty$.
Surprisingly, the 3rd function does not converge uniformly on $[0, 1$. To see this, $f_n(x) = \sqrt{n}xe^{-nx^2}$ has $f_n'(x) = \sqrt{n}e^{-nx^2}\cdot \left(1 - 2nx^2\right) = 0 \iff x = \dfrac{1}{\sqrt{2n}}$. So $\displaystyle \sup_{x \in [0,1]}|f_n(x) - 0| = f_n\left(\frac{1}{\sqrt{2n}}\right) = \dfrac{1}{\sqrt{2e}}$. So $\displaystyle \lim_{n \to \infty} \sup_{x \in [0,1]}|f_n(x) - 0| = \dfrac{1}{\sqrt{2e}} > 0$, proving this function not converging uniformly to $0$ on $[0,1]$.
