Question

Calcule o limite:
$\lim_{x \to 3} (\sqrt{3} - \sqrt{x})/(x^2-2x-3)$

Answer 1

espero ter ajudado...

Answer 2

Bom dia

$\lim_{x \to \ 3 } \dfrac{ \sqrt{3}- \sqrt{x} }{ x^{2} -2x-3}= \lim_{x \to \ 3 } \dfrac{ \sqrt{3}- \sqrt{x} }{ (x-3)(x+1)} }= \\ \\ \lim_{x \to \ 3 } \dfrac{ \sqrt{3}- \sqrt{x} }{ ( \sqrt{x} - \sqrt{3} )( \sqrt{x} + \sqrt{3} )(x+1)} }= \\ \\ \lim_{x \to \ 3 } \dfrac{ \sqrt{3}- \sqrt{x} }{- ( \sqrt{3} - \sqrt{x} )( \sqrt{x} + \sqrt{3} )(x+1)} }= \\ \\ \lim_{x \to \ 3 } \dfrac{ -1 }{( \sqrt{x} + \sqrt{3} )(x+1)} }$

Passando ao limite

$\lim_{x \to \ 3} \dfrac{ \sqrt{3}- \sqrt{x} }{ x^{2} -2x-3} = \dfrac{-1}{( \sqrt{3}+ \sqrt{3} )(3+1)} = \\ \\ \dfrac{-1}{2 \sqrt{3}*4 } = \dfrac{-1}{8 \sqrt{3} }$=

$\dfrac{-1 \sqrt{3} }{8 \sqrt{3} \sqrt{3} } } = \dfrac{- \sqrt{3} }{8*3}=- \dfrac{ \sqrt{3} }{24}$