Question

1)

$lim _{x.. > 2} \frac{x^{4} - 16 }{8 - x^{3} }$

Answer 1

$\lim_{x \to \ 2} \frac{ x^{4}-16 }{8- x^{3} } =[ \frac{0}{0} ]$

$\lim_{x \to \ 2} \frac{ (x^{2} )^{2}-4^{2} }{ 2^{3}- x^{3} }$

$\lim_{x \to \ 2} \frac{( x^{2} -4)( x^{2} +4)}{(2-x)( x^{2}+2x+2^{2})}$

$\lim_{x \to \ 2} \frac{(x-2)(x+2)( x^{2} +4)}{(2-x)( x^{2}+2x+2^{2})}$

$\lim_{x \to \ 2} - \frac{(x+2)( x^{2} +4)}{ x^{2} +2x+4}$

$\lim_{x \to \ 2} - \frac{(2+2)( 2^{2} +4)}{ 2^{2} +2*2+4}$

$\lim_{x \to \ 2} - \frac{4* 8}{4 +4+4}$

$\lim_{x \to \ 2} - \frac{32}{12}$

$\lim_{x \to \ 2} = - \frac{8}{3}$