Question

Suppose that f'(4) = 3 , g'(4) = 7 , g(4) = 4 and g(x) not= to 4 for xnot= to 4 . then cmpute lim xgose to 4 f(g(x))/x-4-f(4)/x-4

Answer 1

It looks like the limit you want to compute is

$\displaystyle \lim_{x\to4} \frac{f(g(x)) - f(4)}{x-4}$

Since $g(4)=4$, this limit corresponds exactly to the derivative of $(f\circ g)(x) = f(g(x))$ at $x=4$. Recall that

$f'(a) = \displaystyle \lim_{x\to a} \frac{f(x) - f(a)}{x - a}$

By the chain rule,

$(f\circ g)'(4) = f'(g(4)) \times g'(4) = f'(4) \times g'(4) = 3\times7 = \boxed{21}$

Since $g'(4)$ exists, $g$ is differentiable at $x=4$ so it must be continuous.