lim_(x→−1) (((√(1+(√(x+5))))−(√3))/(x+1)) =?


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Question Number 119366 by bobhans last updated on 24/Oct/20

$\lim_{x \to - 1} \frac{\sqrt{1 + \sqrt{x + 5}} - \sqrt{3}}{x + 1} = ?$
	Answered by bobhans last updated on 24/Oct/20

	Answered by mathmax by abdo last updated on 24/Oct/20

	$L (x) = \frac{\sqrt{1 + \sqrt{x + 5}} - \sqrt{3}}{x + 1} we do the changement x + 1 = t \Rightarrow$ $L (x) = L (t - 1) = \frac{\sqrt{1 + \sqrt{t + 4}} - \sqrt{3}}{t} (x \to - 1 \Leftrightarrow t \to 0)$ $\Rightarrow L (t - 1) = \frac{\sqrt{1 + \sqrt{t + 4}} - \sqrt{3}) (\sqrt{1 + \sqrt{t + 4}} + \sqrt{3})}{t (\sqrt{1 + \sqrt{t + 4}} + \sqrt{3})}$ $= \frac{1 + \sqrt{t + 4} - 3}{t (\sqrt{1 + \sqrt{t + 4} + \sqrt{3})}} = \frac{(\sqrt{t + 4} - 2) (\sqrt{t + 4} + 2)}{t (\sqrt{1 + \sqrt{t + 4}} + \sqrt{3}) (\sqrt{t + 4} + 2)}$ $= \frac{t}{t (\sqrt{1 + \sqrt{t + 4}} + \sqrt{3}) (\sqrt{t + 4} + 2)} \Rightarrow$ $\lim_{t \to 0} L (t - 1) = \lim_{t \to 0} \frac{1}{(\sqrt{1 + \sqrt{t + 4}} + \sqrt{3}) (\sqrt{t + 4} + 2)}$ $= \frac{1}{2 \sqrt{3} (4)} = \frac{1}{8 \sqrt{3}} = \frac{\sqrt{3}}{24} \Rightarrow \lim_{t \to - 1} L (x) = \frac{\sqrt{3}}{24}$
	Answered by 1549442205PVT last updated on 24/Oct/20

	$Put x + 1 = t \Rightarrow (x \to - 1 \sim t \to 0)$ $\lim_{x \to - 1} \frac{\sqrt{1 + \sqrt{x + 5}} - \sqrt{3}}{x + 1} = \lim_{t \to 0} \frac{\sqrt{1 + \sqrt{t + 4}} - \sqrt{3}}{t}$ $= \lim_{t \to 0} \frac{\sqrt{t + 4} - 2}{t (\sqrt{1 + \sqrt{t + 4}} + \sqrt{3})} = \lim_{t \to 0} \frac{1}{\sqrt{1 + \sqrt{t + 4}} + \sqrt{3}) (\sqrt{t + 4} + 2)}$ $= \frac{1}{2 \sqrt{3} . 4} = \frac{1}{8 \sqrt{3}}$
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