Rewrite [tex]f[/tex] as
[tex]f(x)=\dfrac{10}x=-\dfrac5{1-\frac{x+2}2}[/tex]
and recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n[/tex]
so that for [tex]\left|\dfrac{x+2}2\right|<1[/tex], or [tex]|x+2|<2[/tex],
[tex]f(x)=-5\displaystyle\sum_{n=0}^\infty\left(\frac{x+2}2\right)^n[/tex]
Then the radius of convergence is 2.