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

5/4 -4/5 4/5 none of these

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