We have by distributivity of limits over sums, and continuity of [tex]f(x)[/tex] and [tex]g(x)[/tex], that
[tex]\displaystyle \lim_{x\to0} \bigg(2f(x) - g(x)\bigg) = 2 \lim_{x\to0} f(x) - \lim_{x\to0} g(x) = 2 f(0) - g(0)[/tex]
Given [tex]f(0)=2[/tex], it follows that
[tex]2^2 - g(0) = -3 \implies g(0) = \boxed{7}[/tex]
