If $f(x)=\sqrt{x^{2}+9}$, write the value of $\lim _{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}$.
Given: $f(x)=\sqrt{x^{2}+9}$
Now,
$f(4)=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
So, $\frac{f(x)-f(4)}{x-4}=\frac{\sqrt{x^{2}+9}-5}{x-4}$
On rationalising the numerator, we get
$\frac{f(x)-f(4)}{x-4}=\frac{\sqrt{x^{2}+9}-5}{x-4} \times \frac{\sqrt{x^{2}+9}+5}{\sqrt{x^{2}+9}+5}$
$=\frac{x^{2}+9-25}{(x-4)\left(\sqrt{x^{2}+9}+5\right)}$
$=\frac{x^{2}-16}{(x-4)\left(\sqrt{x^{2}+9}+5\right)}$
$=\frac{(x+4)}{\sqrt{x^{2}+9}+5}$
Taking limit $x \rightarrow 4$, we have
$\lim _{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}=\lim _{x \rightarrow 4} \frac{(x+4)}{\sqrt{x^{2}+9}+5}$
$=\frac{8}{10}$
$=\frac{4}{5}$
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