$\varepsilon-\delta$ definition I am trying to prove the continuity of $f \colon [0, \infty) \to \mathbb{R}$
$$f(x) = \frac{x^2}{x+1}.$$
I tried to use $|f(x) - f(x_0)|$ So:
$$\left|\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}\right|$$
How to get to $|x - x_0|$ so that I can use $|f(x) - f(x_0)| < \delta$?

Consider
$$
\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}=
\frac{x^2x_0+x^2-xx_0^2-x_0^2}{(x+1)(x_0+1)}=
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}(x-x_0)
$$
Thus, if $|x-x_0|<\delta$, you have
$$
\left|\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}\right|<
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}\delta
$$
If you can find an upper bound for the factor to $\delta$, say
$$
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}<M
$$
you can take $\delta=\varepsilon/M$ and then
$$
\left|\frac{x^2}{x+1} - \frac{x_0^2}{x_0+1}\right|<
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}\delta<M\frac{\varepsilon}{M}=\varepsilon
$$
whenever $|x-x_0|<\delta$.
Now
$$
\frac{xx_0+x+x_0}{(x+1)(x_0+1)}=
\frac{xx_0+x+x_0+1-1}{(x+1)(x_0+1)}=
1-\frac{1}{(x+1)(x_0+1)}<1
$$
so you can use $M=1$.
