Question

Evaluate
$\rm \displaystyle \lim_{n\to \sqrt{3}}\ \frac{x^{4}-9}{x^{2}+4\sqrt{3}x-15}$

Answer 1

We know that , denominator can be factorised as
$x^{2}+4\sqrt{3}x-15 \\ \\ = > {x}^{2} + 5 \sqrt{3}x - \sqrt{3}x - 15 \\ \\ = > x(x + 5 \sqrt{3}) - \sqrt{3} (x + 5 \sqrt{3}) \\ \\ = > \: (x + 5 \sqrt{3} )(x - \sqrt{3})$
and numerator can be factorised as

${x}^{4} - 9 \\ \\ = ({ {x}^{2} })^{2} - ( {3)}^{2} \\ \\ = ( {x}^{2} + 3)( {x}^{2} - 3) \\ \\ = ( {x}^{2} + 3)( {x} - \sqrt{3} )(x + \sqrt{3} )$
So

$\lim_{n\to \sqrt{3}}\ \frac{x^{4}-9}{x^{2}+4\sqrt{3}x-15} \\ \\ = \lim_{n\to \sqrt{3}}\ \frac{ ( {x}^{2} + 3)( {x} + \sqrt{3} )(x - \sqrt{3} )}{(x + 5 \sqrt{3} )(x - \sqrt{3})} \\ \\ = \lim_{n\to \sqrt{3}}\ \frac{ ( {x}^{2} + 3)( {x} + \sqrt{3} )}{(x + 5 \sqrt{3} )} \\ \\ apply \: limit \\ \\ = \frac{ ( ({ \sqrt{3} })^{2} + 3)( { \sqrt{3} } + \sqrt{3} )}{( \sqrt{3} + 5 \sqrt{3} )} \\ \\ = \frac{6 \times 2 \sqrt{3} }{6 \sqrt{3} } \\ \\ \lim_{n\to \sqrt{3}}\ \frac{x^{4}-9}{x^{2}+4\sqrt{3}x-15} = 2$
Hope it helps you.