Question

Find the following limit:
$\underset{x\to 9}{lim}(\frac{3-\sqrt{x}}{9-x}+\frac{1}{3-\sqrt{x}}-6.\frac{x^2+162}{729-x^3}).$

Answer 1

$\underset{x\to 9}{lim}(\frac{1}{3+\sqrt{x}}+\frac{1}{3-\sqrt{x}}+\frac{-6x^2-972}{729-x^3})$

$\underset{x\to 9}{lim}(\frac{6}{9-x}+\frac{-6x^2-972}{729-x^3})$

$\underset{x\to 9}{lim}\left(\frac{6(x^2+9x+81)-6x^2-972}{729-x^3}\right)$

$\underset{x\to 9}{lim}\left(\frac{54x-484}{729-x^3}\right)$

Applying L-Hospitals rule and differentiating both numerator and denominator.

$\underset{x\to 9}{lim}\frac{\frac{d}{dx}(54x-484)}{\frac{d}{dx}(729-x^3)}=\underset{x\to 9}{lim}\frac{54}{-3x^2}=-\frac{2}{9}$