Compute, $\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-3x}{x-\sqrt{x+1}-1}$

Solution: Let $x={{\tan }^{2}}\theta $ as $x\to 3\Rightarrow \tan \theta \to \sqrt{3}$

So $\underset{x\to 3}{\mathop{\lim }}\,\frac{{{x}^{2}}-3x}{x-\sqrt{x+1}-}=\underset{\tan \theta \to \sqrt{3}}{\mathop{\lim }}\,\frac{{{\tan }^{2}}\theta \left( {{\tan }^{2}}\theta -3 \right)}{{{\tan }^{2}}\theta -\sqrt{{{\tan }^{2}}\theta +1}-1}$

 but   ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $

Hence $\underset{\tan \theta \to \sqrt{3}}{\mathop{\lim }}\,\frac{{{\tan }^{2}}\theta \left( {{\tan }^{2}}\theta -3 \right)}{{{\tan }^{2}}\theta -\sqrt{{{\tan }^{2}}\theta +1}-1}=\underset{\tan \theta \to \sqrt{3}}{\mathop{\lim }}\,\frac{\left( {{\sec }^{2}}\theta -1 \right)\left( {{\sec }^{2}}\theta -1-3 \right)}{{{\sec }^{2}}\theta -1-\sec \theta -1}$ 

$=\underset{\tan \theta \to \sqrt{3}}{\mathop{\lim }}\,\frac{\left( {{\sec }^{2}}\theta -1 \right)\left( {{\sec }^{2}}\theta -4 \right)}{{{\sec }^{2}}\theta -\sec \theta -2}=\underset{\tan \theta \to \sqrt{3}}{\mathop{\lim }}\,\frac{\left( {{\sec }^{2}}\theta -1 \right)\left( \sec \theta -2 \right)\left( \sec \theta +2 \right)}{\left( \sec \theta +1 \right)\left( \sec \theta -2 \right)}$

$=\underset{\tan \theta \to \sqrt{3}}{\mathop{\lim }}\,\frac{\left( \sec \theta -1 \right)\left( \sec \theta +1 \right)\left( \sec \theta +2 \right)}{\sec \theta +1}=\underset{\sec \theta \to 2}{\mathop{\lim }}\,\left( \sec \theta -1 \right)\left( \sec \theta +2 \right)=4$