Solution: Given that the function is \[ f(x) = \frac{x^3}{1 + x^2} = x - \frac{x}{1 + x^2}, \quad x \in \mathbb{R} . \] We need to show that $f$ is continuous on $\mathbb{R} $. Let $a \in \mathbb{R} $. We claim that $f$ is continuous at $a$. Recall that a function $f$ is continuous at $x$ if any sequence $\left( x_n \right) $ converging to $x$, the sequence $\left( f(x_n) \right) $ will converge to $f(x)$.

Let $\left( a_n \right) $ be a sequence converging to $a$. Since $x \mapsto x^3$ and $x \mapsto 1 + x^2$ are polynomials and hence continuous, so we must have \[ \lim_{n \to \infty} a_n^3 = a^3 \quad \text{ and } \quad \lim_{n \to \infty} \left( 1 + a_n^2 \right) = 1 + a^2. \] As $1 + a_n^2 > 0$ for any $n \in \mathbb{N} $ and also $1 + a^2 > 0$, the limit \[ \lim_{n \to \infty} \frac{a_n^3}{1 + a_n^2} = \frac{a^3}{1 + a^2} = f(a). \] Thus, $f$ is continuous at $a$ and hence $f$ is continuous on $\mathbb{R}$.

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Let us show the continuity by the $\epsilon $-$\delta $ method. Note that for any $x,y \in \mathbb{R} $, we have \begin{align*} \left\vert f(x) - f(y) \right\vert & = \left\vert \left( x - \frac{x}{1 + x^2} \right) - \left( y - \frac{y}{1 + y^2} \right) \right\vert \\ & = \left\vert x - y - \frac{x}{1 + x^2} + \frac{y}{1 + y^2} \right\vert \\ & \leq \vert x - y \vert + \left\vert \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right\vert. \end{align*} We will now use the inequality $2 \vert xy \vert \leq x^2 + y^2$ (AM-GM inequality) to estimate the last quantity in the above inequality. \begin{align*} \left\vert \frac{x}{1 + x^2} - \frac{y}{1 + y^2} \right\vert & = \left\vert \frac{x + xy^2 - y - x^2 y}{(1 + x^2)(1 + y^2)} \right\vert \\ & = \left\vert \frac{(x-y)( 1 - xy)}{(1 + x^2) (1 + y^2)} \right\vert \\ & \leq \vert x - y \vert \cdot \left( \frac{1 + \vert xy \vert}{(1 + x^2)(1 + y^2)} \right) \\ & \leq \vert x - y \vert \cdot \left( \frac{1 + \frac{x^2 + y^2}{2}}{(1 + x^2)(1 + y^2)} \right) \\ & = \frac{\vert x - y \vert }{2} \left( \frac{1 + x^2 + 1 + y^2}{(1 + x^2) (1 + y^2)} \right) \\ & \leq \frac{\vert x - y \vert }{2} \left( \frac{1}{1 + x^2} + \frac{1}{1 + y^2} \right) \\ \leq \vert x - y \vert. \end{align*} Note that the last inequality holds because $\frac{1}{1 + x^2} + \frac{1}{1 + y^2} \leq 2$. Thus, we showed that \[ \vert f(x) - f(y) \vert \leq 2 \vert x - y \vert . \] Thus $f$ is Lipschitz and hence continuous and uniform continuos on $\mathbb{R} $. (We could have also take $\delta = \frac{\epsilon }{2}$ to show the continuity and uniform continuity).