Question

Evaluate the limit:

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

Answer 1

At $x\to 7,\frac{4-\sqrt{9+x}}{1-\sqrt{8-x}}\text{becomes}\frac{0}{0}form$

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

On rationalising numerator and denominator, we get

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

$=\underset{x\to 7}{lim}\frac{(4^2-{\left(\sqrt{9+x}\right)}^2)(1+\sqrt{8-x})}{(1^2-{\left(\sqrt{8-x}\right)}^2)(4+\sqrt{9+x})}$

$[\because (a-b\left)\right(a+b)=a^2-b^2]$

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

$=\underset{x\to 7}{lim}\frac{-(x-7)(1+\sqrt{8-x})}{(x-7)(4+\sqrt{9+x})}\left[\right(x-7)\ne 0]$

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

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

$=\frac{-(1+1)}{(4+4)}$

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