It looks like the limit you want to compute is
[tex]\displaystyle \lim_{x\to4} \frac{f(g(x)) - f(4)}{x-4}[/tex]
Since [tex]g(4)=4[/tex], this limit corresponds exactly to the derivative of [tex](f\circ g)(x) = f(g(x))[/tex] at [tex]x=4[/tex]. Recall that
[tex]f'(a) = \displaystyle \lim_{x\to a} \frac{f(x) - f(a)}{x - a}[/tex]
By the chain rule,
[tex](f\circ g)'(4) = f'(g(4)) \times g'(4) = f'(4) \times g'(4) = 3\times7 = \boxed{21}[/tex]
Since [tex]g'(4)[/tex] exists, [tex]g[/tex] is differentiable at [tex]x=4[/tex] so it must be continuous.
