If [tex]f[/tex] and [tex]g[/tex] are continuous everywhere, that means that for any [tex]x=c[/tex], the limit as [tex]x\to c[/tex] for either function is the value of that function at [tex]c[/tex]:
[tex]\displaystyle\lim_{x\to c}f(x)=f(c)\,\text{and}\,\lim_{x\to c}g(x)=g(c)[/tex]
Applying some properties of limits, we can rewrite the original limit as
[tex]\displaystyle\lim_{x\to2}(f(x)+4g(x))=\lim_{x\to2}f(x)+\lim_{x\to2}4g(x)=\lim_{x\to2}f(x)+4\lim_{x\to2}g(x)=16[/tex]
Given the continuity of [tex]f[/tex] and [tex]g[/tex], we have
[tex]f(2)+4g(2)=16\implies 4g(2)=16-3=13\implies g(2)=\dfrac{13}4[/tex]
So for both parts, the answer is 13/4.
