Solve this
Question:

If $f(x)=\sqrt{x^{2}+9}$, write the value of $\lim _{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}$.

Solution:

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

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

$=\frac{8}{10}$

$=\frac{4}{5}$