Check for the below the pointwise and uniform convergence.
$\displaystyle{f_n(x)=\frac{x\sqrt{n}}{1+nx^2}}, \ x\in \mathbb{R}$
$\displaystyle{g_n(x)=n^2xe^{-nx}}, \ x\in \mathbb{R}^+$
$\displaystyle{h_n(x)=\frac{\sin nx}{1+nx^2}}, \ x\in \mathbb{R}$
For the pointwise convergence we have to calculate the limit as $n\rightarrow +\infty$ and if we get a function $f\neq \infty$ then it converges, right?
we have the following:
For $x\neq 0$ we have that $$\lim_{n\rightarrow +\infty}f_n(x)=\lim_{n\rightarrow +\infty}\frac{x\sqrt{n}}{1+nx^2}=\lim_{n\rightarrow +\infty}\frac{n\left (xn^{-\frac{1}{2}}\right )}{n\left (\frac{1}{n}+x^2\right )}=\lim_{n\rightarrow +\infty}\frac{xn^{-\frac{1}{2}}}{\frac{1}{n}+x^2}=\frac{0}{0+x^2}=0$$ For $x= 0$ we have that $$\lim_{n\rightarrow +\infty}f_n(0)=\lim_{n\rightarrow +\infty}\frac{0}{1}=\lim_{n\rightarrow +\infty}0=0$$ Therefore $f_n(x)$ converges pointwise, right?
For $x\neq 0$ we have that $$\lim_{n\rightarrow +\infty}g_n(x)=\lim_{n\rightarrow +\infty}n^2xe^{-nx}\overset{\text{De L'Hopital}}{=}\lim_{n\rightarrow +\infty}\frac{2n}{x^2e^{nx}}\overset{\text{De L'Hopital}}{=}\lim_{n\rightarrow +\infty}\frac{2}{x^3e^{nx}}=0$$ For $x= 0$ we have that $$\lim_{n\rightarrow +\infty}g_n(0)=\lim_{n\rightarrow +\infty}0=0$$ Therefore $g_n(x)$ converges pointwise, right?
For $x\neq 0$ we have that $$\lim_{n\rightarrow +\infty}h_n(x)=\lim_{n\rightarrow +\infty}\frac{\sin nx}{1+nx^2}$$ We have that \begin{align*}-1\leq \sin nx\leq 1&\Rightarrow -\frac{1}{1+nx^2}\leq \frac{\sin nx}{1+nx^2}\leq \frac{1}{1+nx^2}\\ & \Rightarrow -\lim_{n\rightarrow +\infty}\frac{1}{1+nx^2}\leq \lim_{n\rightarrow +\infty}\frac{\sin nx}{1+nx^2}\leq \lim_{n\rightarrow +\infty}\frac{1}{1+nx^2} \\ & \Rightarrow -0\leq \lim_{n\rightarrow +\infty}\frac{\sin nx}{1+nx^2}\leq 0\end{align*} Therefore $\lim_{n\rightarrow +\infty}\frac{\sin nx}{1+nx^2}=0$ For $x= 0$ we have that $$\lim_{n\rightarrow +\infty}h_n(0)=\lim_{n\rightarrow +\infty}\frac{0}{1}=0$$ So $h_n(x)$ converges pointwise, right?
How can we check the uniform convergence?
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EDIT:
Do we maybe do the following for the uniform convergence?
\begin{align*}&f_n(x)=\frac{x\sqrt{n}}{1+nx^2}\\ &\rightarrow f_n'(x)=\frac{\sqrt{n}(1+nx^2)-x\sqrt{n}2nx}{(1+nx^2)^2}=\frac{\sqrt{n}-n\sqrt{n}x^2}{(1+nx^2)^2} =\frac{\sqrt{n}\left (1-nx^2\right )}{(1+nx^2)^2} \\ & \rightarrow f'_n(x)=0 \Rightarrow \sqrt{n}\left (1-nx^2\right )=0 \Rightarrow 1-nx^2=0 \Rightarrow x^2=\frac{1}{n}\Rightarrow x=\pm \frac{1}{\sqrt{n}} \\ & f_n\left (-\frac{1}{\sqrt{n}}\right )=\frac{-1}{2} \\ & f_n\left (\frac{1}{\sqrt{n}}\right )=\frac{1}{2}\end{align*} So the maximum is $\frac{1}{2}$. For $n\rightarrow +\infty$ the limit is $\frac{1}{2}\neq 0$, so this sequence does not converge uniformly. Is this correct?
\begin{align*}&g_n(x)=n^2xe^{-nx}\\ &\rightarrow g_n'(x)=n^2e^{-nx}-n^3xe^{-nx}=\left (n^2-n^3x\right )e^{-nx} \\ &\rightarrow g_n'(x)=0 \Rightarrow \left (n^2-n^3x\right )e^{-nx}=0 \Rightarrow n^2-n^3x=0 \Rightarrow \frac{1}{n} \\ &g_n\left (\frac{1}{n}\right )=\frac{n}{e}\end{align*} So the maximum is $\frac{n}{e}$. For $n\rightarrow +\infty$ the limit is $+\infty\neq 0$, so this sequence does not converge uniformly. Is this correct?
\begin{align*}|h_n(x)|=\left |\frac{\sin nx}{1+nx^2} \right |\leq \frac{ nx}{1+nx^2}\end{align*} How do we continue?
Most easy is to use necessary and sufficient condition $$\lim\limits_{n \to \infty}\sup|f_n(x)-f(x)| =0$$ and for $\sup$ is possible to use derivative, for example.
Let's take for 1st one $f^{'}(x)=\frac{\sqrt{n}(1-nx^2)}{(1+nx^2)^2}$. These gives extremums for $f$ in $\pm \frac{1}{\sqrt{n}}$ so maximum will be $\frac{1}{2}$. Of course derivative is only one, but powerful, tool for finding $\sup$.