Q1 We have $f_n(x) = \frac{x}{1+nx^{2}} = \frac{1}{\sqrt{n}} \frac{\sqrt{n}x}{1+nx^{2}} = \frac{1}{\sqrt{n}} \frac{t}{1+t^{2}} $where $t=\sqrt{n}x.$ Since $|t|/(1+t^2)\leq 1,$ we get $\forall x\in \mathbb{R}: \quad |f_n(x)-0|\leq \frac{1}{\sqrt{n}}.$It follows from the sandwich rule that $\lVert f_n - 0 \rVert_{\infty} \to 0$ as $n\to \infty$, that is, $f_n \rightrightarrows 0$, and as a result $f_n \to 0.$ Using the quotient rule for derivatives, $f_{n}'(x) = \frac{(1+ nx^2)-2nx^2}{(1+nx^2)^2} = \frac{1-nx^2}{(1+nx^2)^2}$hence $f'(n) \; \substack{\longrightarrow \\ n \to \infty} \; \begin{cases} 1 & x=0 \\ 0 & x \neq 0\end{cases}$The convergence is not uniform as the limit is not continuous while each $f_{n}'$ is. Q2 In order to show that $f_n$ does not converge uniformly to $f$ in $\Omega,$ we need to show that there exists $\varepsilon >0$ such that for all positive integers $N,$ there exists $n>N$ and $x\in \Omega$ such that $\lvert f_n(x)-f(x) \rvert \geq \varepsilon$By definition, $\lvert f_n(x_n)-f(x_n)\rvert \not \to 0$ means that for all $N\in \mathbb{N},$ we can indeed choose such $n>N$ and $x:=x_n$ such that $\lvert f_n(x_n)-f(x_n)\rvert \geq \varepsilon.$ Therefore, $\lvert f_n(x_n)-f(x_n)\rvert \not \to 0 \implies f_n \not \rightrightarrows f.$ $\square\newline$ We define $f_n:(0,1) \to \mathbb{R}$ by $f_n(x)=x^n.$ For all $x\in(0,1),$ we have $x^n \to 0$ so $f_n \to 0.$ Hence, it is sufficient to show that $f_n$ does not converge uniformly to $f=0. \newline$ Let $x_1 = 0.1$ and $x_{n} = 1-\frac{1}{n}$ for all $n > 1.$ Then each $x_n$ lies in $(0,1).$ For all $n>1,$ we have $f_n(x_n) = \left(1-\frac{1}{n}\right)^n \to \frac{1}{e}, \quad \text{as } n\to \infty $(since $e^x = \lim_{n \to \infty} (1+x/n)^n$). That is, $|f_n(x_n)-0| \not \to 0$ which implies, by the above theorem, that $f_n \not \rightrightarrows 0.$ $\square$ Q3 The statement is clearly true when $x=y.$ Suppose otherwise that $x \neq y.$ Since $f_n \to f,$ we can choose a positive integer $N$ such that for all $x\in [a,b],$ $|f_N(x) - f(x)| < \frac{|x-y|}{2}.$ Applying the triangle inequality, we get$\begin{align*}|f(x)-f(y)| & \leq |f(x)-f_N(x)|+|f_N(x)-f_N(y)|+|f_N(y) - f(y)| \\ & \leq (K+1) |x-y|.\end{align*}$ $\square \newline$ Q4 For any fixed $x>0,$ $\frac{2nx}{1+n^3x^2}\leq \frac{2nx}{n^3x^2} = \frac{2}{n^2x}\to 0, \quad \text{as } n \to \infty.$Also $g_{n}(0)=0$ so $\lVert g_{n}\rVert_{\infty} \to 0,$ as $n \to \infty$ (that is, $g_{n}\rightrightarrows 0$). Notice that for all $x\in [0,1],$ $\log(g_{n}(x))\leq g_{n}(x)$ so $\log(g_{n})\to 0.$ Also, $\log(1+n^3x^2) =\log(g_{n}(x))-\log(2nx).$ So $f_{n}(x)=\frac{\log(1+n^3x^2)}{n^2} = \frac{\log(g_{n}(x))}{n^2} - \frac{\log(2nx)}{n^2} \to 0$(since $\log (2nx)\leq 2nx$ implies $\log(2nx)/n^2 \leq \frac{2nx}{n^2} \to 0$). Therefore $f_{n}\rightrightarrows 0.$