Question

$\underset{x\to 7}{\lim }\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}$

Answer 1

$\underset{x\to 7}{\lim }\left[\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}\right]$

It is of the form $\frac{0}{0}$.
Rationalising the numerator and the denominator:

$\underset{x\to 7}{\lim }\left[\frac{\left(4-\sqrt{9+x}\right)}{1}\times \frac{\left(4+\sqrt{9+x}\right)}{\left(4+\sqrt{9+x}\right)}\times \frac{1}{\left(1-\sqrt{8-x}\right)}\times \frac{\left(1+\sqrt{8-x}\right)}{\left(1+\sqrt{8-x}\right)}\right]$

= $\underset{x\to 7}{\lim }\left[\frac{16-\left(9+x\right)}{\left(4+\sqrt{9+x}\right)}\times \frac{\left(1+\sqrt{8-x}\right)}{1-\left(8-x\right)}\right]$

= $\underset{x\to 7}{\lim }\left[\frac{-1\left(-7+x\right)\left(1+\sqrt{8-x}\right)}{\left(4+\sqrt{9+x}\right)\left(-7+x\right)}\right]$

= $\underset{x\to 7}{\lim }\left[\frac{-\left(1+\sqrt{8-x}\right)}{4+\sqrt{9+x}}\right]$

= $-\left(\frac{1+\sqrt{8-7}}{4+\sqrt{9+7}}\right)$

= $\frac{-2}{4+4}$

= $\frac{-1}{4}$