If $f(x)=\sqrt{x^2+9}$, then $\lim\limits_{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}$ has the value

4/5

We have,

$f(x) =\sqrt{x^2+9}$

$\Rightarrow f'(x) =\frac{x}{\sqrt{x^2+9}}$

Now,

$\lim\limits_{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}=f'(4)$                     [By def. of derivative]

$\Rightarrow \lim\limits_{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}=\frac{4}{\sqrt{4^2+9}}=\frac{4}{5}$