Question

Need help solving this limit problem.

Answer 1

If $f$ and $g$ are continuous everywhere, that means that for any $x=c$, the limit as $x\to c$ for either function is the value of that function at $c$:

$\displaystyle\lim_{x\to c}f(x)=f(c)\,\text{and}\,\lim_{x\to c}g(x)=g(c)$

Applying some properties of limits, we can rewrite the original limit as

$\displaystyle\lim_{x\to2}(f(x)+4g(x))=\lim_{x\to2}f(x)+\lim_{x\to2}4g(x)=\lim_{x\to2}f(x)+4\lim_{x\to2}g(x)=16$

Given the continuity of $f$ and $g$, we have

$f(2)+4g(2)=16\implies 4g(2)=16-3=13\implies g(2)=\dfrac{13}4$

So for both parts, the answer is 13/4.