Question

resolva o limite
$\lim_{x \to \ 4} \frac{\sqrt{16-x^{2} } }{x-4}$

Answer 1

$\lim_{x \to \ 4} \frac{\sqrt{16-x^{2} } }{x-4} \\ \lim_{x \to \ 4} \frac{ (\sqrt{16 - {x}^{2} })( \sqrt{16- {x}^{2} } )}{(x - 4)( \sqrt{16-{x}^{2} )} }$

$\lim_{x \to \ 4} \frac{ { (\sqrt{16 - {x}^{2} })}^{2} }{(x - 4)( \sqrt{16-{x}^{2} }) } \\ \lim_{x \to \ 4} \frac{16 - {x}^{2} }{(x - 4)( \sqrt{16-{x}^{2}) } }$

$\lim_{x \to \ 4} - \frac{(4 + x)(x - 4)}{(x - 4)( \sqrt{16-{x}^{2} }) } \\\lim_{x \to \ 4} -\frac{4 + x}{ \sqrt{16-{x}^{2} } } = -\frac{4+4}{ \sqrt{16-{4}^{2} } } \\ = -\frac{8}{ \sqrt{0} } =-∞$