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Compute: $\displaystyle\lim_{x\rightarrow 2}\frac{\sqrt{x+1}-\sqrt{ 1-x}}{x}$

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sqrt x+1 means $\sqrt{x} + 1$ or $\sqrt{x+1}$? What about sqrt 1 - x? I'm guessing $\sqrt{1-x}$ –  Pragabhava Oct 25 '12 at 0:07
    
$(\sqrt3-i)/2$. –  Berci Oct 25 '12 at 0:09
    
    
L'Hopital is not a good way of learning what the limits are but a really good way to check your answer. –  GYC Oct 25 '12 at 0:13
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I'm not sure that the problem was to compute the limit when x --> 2... –  Provost Oct 25 '12 at 0:18

2 Answers 2

I assume you are interested in $$\lim_{x \to 0} \dfrac{\sqrt{1+x} - \sqrt{1-x}}x$$ Multiply and divide by $\sqrt{1+x} + \sqrt{1-x}$ to get $$\dfrac{\sqrt{1+x} - \sqrt{1-x}}x \times \dfrac{\sqrt{1+x} + \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}} = \dfrac{(1+x)-(1-x)}{x(\sqrt{1+x} + \sqrt{1-x})} = \dfrac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}\\ = \dfrac2{\sqrt{1+x} + \sqrt{1-x}}$$ Now can you finish it off?

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Just evaluate the function at $x = 2$.

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