Since [tex]f[/tex] is continuous, you know that
[tex]\displaystyle\lim_{x\to3}f(x)=f(3)=5[/tex]
So,
[tex]\displaystyle\lim_{x\to3}(2f(x)-g(x))=2f(3)-\lim_{x\to3}g(x)=10-g(3)=4[/tex]
where the limit for [tex]g[/tex] also is equal to the value of [tex]g(3)[/tex] because [tex]g[/tex] is also continuous. This means [tex]g(3)=6[/tex].
