Solution: We recall that a function $f:U\subseteq \mathbb{R} \to \mathbb{R} $ is said to be uniformly continuous on $U$ if for every $\varepsilon >0$ there exists $\delta >0$ such that for all $x,y\in U$ if $|x-y|< \delta $ then $\vert f(x)-f(y) \vert < \varepsilon $. Equivalently, if for every sequences $\left( x_n \right) $ and $\left( y_n \right) $ if $\left\vert x_n - y_n \right\vert \to 0$ as $n\to \infty $, then $\left\vert f\left( x_n \right) - f\left( y_n \right) \right\vert \to \infty $ as $n\to \infty $.

• Take $x_n = n$ and $y_n = n + \frac{1}{n}$, then \[ \left\vert x_n - y_n \right\vert = \left\vert n - n - \frac{1}{n} \right\vert = \left\vert \frac{1}{n} \right\vert \to 0 \text{ as } n\to \infty . \] Whereas, \begin{align*} \left\vert f\left( x_n \right) - f\left( y_n \right) \right\vert & = \left\vert n^2 - \left( n + \frac{1}{n} \right) ^2 \right\vert \\ & = \left\vert -2 - \frac{2}{n^2} \right\vert \to 2, \end{align*} hence $f$ is not uniformly continuous on $\mathbb{R} $.

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• We consider the sequences, \[ x_n^2 = n\pi,~ y_n^2 = n\pi + \frac{\pi}{2}. \] Note that \begin{align*} \left\vert x_n - y_n \right\vert & = \left\vert \sqrt{n\pi} - \sqrt{n\pi + \frac{\pi}{2}} \right\vert \\ & = \left\vert \left( \sqrt{n\pi} - \sqrt{n\pi + \frac{\pi}{2}} \right) \times \frac{\sqrt{n\pi} + \sqrt{n\pi + \frac{\pi}{2}}}{\sqrt{n\pi} + \sqrt{n\pi + \frac{\pi}{2}}} \right\vert \\ & = \left\vert \frac{n\pi - n\pi - \frac{\pi}{2}}{\sqrt{n\pi} + \sqrt{n\pi + \frac{\pi}{2}}} \right\vert \\ & = \left\vert \frac{-\frac{\pi}{2}}{\sqrt{n\pi} + \sqrt{n\pi + \frac{\pi}{2}} } \right\vert \to 0,~\text{ as } n\to \infty . \end{align*} But \begin{align*} \left\vert g\left( x_n \right) - g\left( y_n \right) \right\vert & = \left\vert \sin x_n^2 - \sin y_n^2 \right\vert \\ & = \left\vert \sin n\pi - \sin \left( n\pi + \frac{\pi}{2} \right) \right\vert \\ & = \left\vert \sin \frac{\pi}{2} \right\vert = 1. \end{align*} Hence, the function is not uniformly continuous on $\mathbb{R} $.

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• We recall that if we can extend a continuous function defined on $(a,b)$ to $[a,b]$ and the extended function is uniformly continuous on $[a,b]$, then $f$ is uniformly continuous on $(a,b)$. Here the function if $f(x) = x \sin \frac{1}{x}$, $x\in(0,1)$. Note that the given function is continuous on $(0,1)$ and \[ \lim_{x \to 0} x \sin \frac{1}{x} = 0, \] as $\sin \frac{1}{x}$ is bounded. Also, \[ \lim_{x \to 1} x \sin \frac{1}{x} = \sin 1. \] So we extended the given function as \[ \tilde{h}(x) = \begin{dcases} x \sin \frac{1}{x}, &\text{ if } x\in (0,1] ;\\ 0, &\text{ if } x = 0. \end{dcases} \] The function $\tilde{h} $ is continuous on $[0,1]$ and hence uniformly continuous on $[0,1]$ (continuous function on a compact set is uniformly continuous). Therefore, the given function is uniformly continuous on $(0,1)$.

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• The function is $w(x) = x^2 \sin x$ is not uniformly continuous. Take the sequences \[ x_n = n\pi ~ y_n = n\pi + \frac{1}{n}. \] It is clear that $\left\vert x_n - y_n \right\vert \to 0$ as $n\to \infty $. Now we consider \begin{align*} \left\vert w\left( x_n \right) - w\left( y_n \right) \right\vert & = \left\vert x_n^2 \sin x_n - y_n^2 \sin y_n \right\vert \\ & = \left\vert n^2\pi^2 \sin (n\pi) - \left( n\pi + \frac{1}{n} \right)^2 \sin \left( n\pi + \frac{1}{n} \right) \right\vert \\ & = \left\vert n^2 \sin \left( \frac{1}{n} \right) + 2\pi \sin \left( \frac{1}{n} \right) + \frac{1}{n^2} \sin \left( \frac{1}{n} \right) \right\vert. \end{align*} But note that \[ n^2 \sin \frac{1}{n} = n \frac{\sin \left( \frac{1}{n} \right) }{\frac{1}{n}} \to \infty,~ \text{as } n \to \infty . \] Thus, $w$ is not uniformly continuous on $\mathbb{R} $.