Question

$\lim_{ \to \0} \sqrt{1+x} -1 / -x$
Quanto x tende a 0 

Answer 1

Resolvendo, temos:

$\lim_{x \to 0} \dfrac{ \sqrt{1 + x} - 1 }{- x} = \lim_{x \to 0} \dfrac{ \sqrt{1 + x} - 1 }{- x} * \dfrac{ \sqrt{1+x} + 1 }{ \sqrt{1+x} + 1 } \\ \\ \\ \lim_{x \to 0} \dfrac{ \sqrt{1 + x} - 1 }{- x} * \dfrac{ \sqrt{1+x} + 1 }{ \sqrt{1+x} + 1 } = \lim_{x \to 0} \dfrac{ (\sqrt{1+x})^2 - 1^2 }{-x*( \sqrt{1+x} + 1 )} \\ \\ \\ \lim_{x \to 0} \dfrac{ (\sqrt{1+x})^2 - 1^2 }{-x*( \sqrt{1+x} + 1 )} = \lim_{x \to 0} \dfrac{1 + x - 1}{-x*( \sqrt{1+x} + 1 )}$

$\lim_{x \to 0} \dfrac{1 + x - 1}{-x*( \sqrt{1+x} + 1 )} = \lim_{x \to 0} \dfrac{x}{-x*( \sqrt{1+x} + 1 )} \\ \\ \\ \lim_{x \to 0} \dfrac{x}{-x*( \sqrt{1+x} + 1 )} = \lim_{x \to 0} \dfrac{1}{-1*( \sqrt{1+x} + 1 )} \\ \\ \\ \lim_{x \to 0} \dfrac{1}{-1*( \sqrt{1+x} + 1 )} = \dfrac{1}{-1*( \sqrt{1+0} +1) } = -\dfrac{1}{2}$