Limit Value=?

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Limit Value=?

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Answer 1 · 2018-03-06T18:29:23

$(2)$ $(2)$

Explanation:

$\frac{\sqrt{3 x} - 3}{\sqrt{2 x - 4} - \sqrt{2}} = \frac{\sqrt{3 x} - 3}{\sqrt{2 x - 4} - \sqrt{2}} (\frac{\sqrt{2 x - 4} + \sqrt{2}}{\sqrt{2 x - 4} + \sqrt{2}}) =$ $(sqrt(3x)-3)/(sqrt(2x-4)-sqrt2) =(sqrt(3x)-3)/(sqrt(2x-4)-sqrt2) ((sqrt(2x-4)+sqrt2)/(sqrt(2x-4)+sqrt2))=$

$= \frac{(\sqrt{3 x} - 3) (\sqrt{2 x - 4} + \sqrt{2})}{2 x - 4 - 2} = \frac{(\sqrt{3 x} - 3) (\sqrt{2 x - 4} + \sqrt{2})}{2 x - 6}$ $=((sqrt(3x)-3)(sqrt(2x-4)+sqrt2))/(2x-4-2) =((sqrt(3x)-3)(sqrt(2x-4)+sqrt2))/(2x-6)$

but

$\frac{\sqrt{3 x} - 3}{2 x - 6} = \frac{\sqrt{3} (\sqrt{x} - \sqrt{3})}{2 ({(\sqrt{x})}^{2} - {(\sqrt{3})}^{2})} = \frac{\sqrt{3}}{2} \frac{1}{\sqrt{x} + \sqrt{3}}$ $(sqrt(3x)-3)/(2x-6) = (sqrt3(sqrtx-sqrt3))/(2((sqrtx)^2-(sqrt3)^2)) = sqrt3/2 1/(sqrtx+sqrt3)$

so

$lim_(x->3)(sqrt(3x)-3)/(sqrt(2x-4)-sqrt2) =lim_(x->3)(sqrt3/2)(sqrt(2x-4)+sqrt2)/(sqrtx+sqrt3) = 1/sqrt2$

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