Question

Evaluate the limit:

$\underset{x\to 5}{lim}\frac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}}$

Answer 1

$\underset{x\to 5}{lim}\frac{x-5}{\sqrt{6x-5}-\sqrt{4x+5}},\left(\frac{0}{0}form\right)$

On rationalising denominator, we get

$\underset{x\to 5}{lim}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{(\sqrt{6x-5}-\sqrt{4x+5})(\sqrt{6x-5}+\sqrt{4x+5})}$

$\underset{x\to 5}{lim}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{({\left(\sqrt{6x-5}\right)}^2-{\left(\sqrt{4x+5}\right)}^2)}$

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

$=\underset{x\to 5}{lim}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{(6x-5-4x-5)}$

$=\underset{x\to 5}{lim}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{(2x-10)}$

$=\underset{x\to 5}{lim}\frac{(x-5)(\sqrt{6x-5}+\sqrt{4x+5})}{2(x-5)}\left[\right(x-5)\ne 0]$

$=\underset{x\to 5}{lim}\frac{(\sqrt{6x-5}+\sqrt{4x+5})}{2}$

$=\frac{(\sqrt{6\left(5\right)-5}+\sqrt{4\left(5\right)+5})}{2}$

$=\frac{(\sqrt{25}+\sqrt{25})}{2}$

$=5$