Proving continuity of the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $

Question

Using only the epsilon delta criterion, show that the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ is continuous over its entire domain of definition.

Definition of the epsilon delta criterion:

A function $f$ from $\Bbb R$ to $\Bbb R$ is continuous at a point $x_0 \in \mathbb{R} $ if given $ε > 0$ there exists $δ > 0$ such that if $|x-x_0| < \delta $ then $|f(x)-f(x_0)|< \epsilon$.

I have already started looking for delta, but get stuck here, and don't know how to further transform or estimate to find the appropriate delta:

$ \begin{aligned}\left|f(x)-f\left(x_{0}\right)\right| &=\left|\frac{x^{3}}{1+x^{2}}-\frac{x_{0}^{3}}{1+x_{0}^{2}}\right|=\\ &=\left|\frac{x^{3}\left(1+x_{0}^{2}\right)}{\left(1+x^{2}\right)\left(1+x_{0}^{2}\right)}-\frac{x_{0}^{3}\left(1+x^{2}\right)}{\left(1+x_{0}^{2}\right)\left(1+x^{2}\right)}\right|=\\ &=\left|\frac{x^{3}\left(1+x_{0}^{2}\right)-x_{0}^{3}\left(1+x^{2}\right)}{\left(1+x^{2}\right)\left(1+x_{0}^{2}\right)}\right|=?\end{aligned} $

Answer 1

We have

\begin{align} |x^3(1+y^2)- y^3(1+x^2)| & = |(x-y)(x^2+x y + y^2 + x^2 y^2)|\\ &\le |x-y|(x^2 + \frac{1}{2}(x^2+y^2) + y^2 + x^2 y^2)\\ &\le \frac{3}{2}|x-y|(x^2 + y^2 + x^2 y^2)\\ &\le \frac{3}{2}|x-y|(1 + x^2)(1+y^2)\\ \end{align} Hence \begin{equation} \frac{|x^3(1+y^2)- y^3(1+x^2)|}{(1 + x^2)(1+y^2)} \le \frac{\frac{3}{2}|x-y|(1 + x^2)(1+y^2)}{(1 + x^2)(1+y^2)} \le\frac{3}{2}|x-y| \end{equation}

Hence \begin{equation} |f(x)-f(x_0)|\le \frac{3}{2} |x-x_0| \end{equation}

Answer 2

We have

$$\frac{x^3}{1+x^2}-\frac{x_0^3}{1+x_0^2}=(x-x_0)\left(\dfrac{xx_0-1}{(1+x^2)(1+x_0^2)}+1\right).$$

Clearly, the right factor is bounded, for any $x_0$, as the denominator has no root and its degree exceeds that of the numerator.

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As $\left|\dfrac x{x^2+1}\right|$ is bounded by $\frac12$ and $\dfrac1{x^2+1}$ by $1$, the factor is bounded by $\dfrac{\frac12|x_0|+1}{1+x_0^2}+1$, itself bounded by $\frac12\frac12+1+1$. Hence, convergence is uniform.