Question

$\frac{ \sqrt{1 + t - \sqrt{1 - t} } }{t}$
t tende a 0​

Answer 1

Temos que

$\displaystyle\lim_{t\to0} \frac{ \sqrt{1 + t} - \sqrt{1 - t} }{t} \\ = \displaystyle\lim_{t\to0} \frac{ \sqrt{1 + t} - \sqrt{1 - t} }{t} \cdot\frac{ \sqrt{1 + t} + \sqrt{1 - t} }{\sqrt{1 + t} + \sqrt{1 - t}} \\ = \displaystyle\lim_{t\to0} \frac{ (\sqrt{1 + t})^{2} + (\sqrt{1 - t})^{2} }{t(\sqrt{1 + t} + \sqrt{1 - t})} \\ = \displaystyle\lim_{t\to0} \frac{(1 + t) - (1 - t) }{t(\sqrt{1 + t} + \sqrt{1 - t})} \\ = \displaystyle\lim_{t\to0} \frac{1 + t - 1 + t }{t(\sqrt{1 + t} + \sqrt{1 - t})} \\ = \displaystyle\lim_{t\to0} \frac{2t }{t(\sqrt{1 + t} + \sqrt{1 - t})} \\ = \displaystyle\lim_{t\to0} \frac{2 }{\sqrt{1 + t} + \sqrt{1 - t}} \\ = \dfrac{2}{\sqrt{1 + 0} + \sqrt{1 - 0} } \\ = \dfrac{2}{\sqrt{1} + \sqrt{1} } = \dfrac{2}{1 + 1} = \dfrac{2}{2} = 1$