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About LinkBacksNot sure how to go about this.
Can I just rationalize either the numerator or denominator?
lim (√2-x) - (√8-4x)
x->2 (√3x+3) - (√11-x)
HiOriginally Posted by Neconine
The numerator is $\displaystyle \sqrt{2-x}-\sqrt{4(2-x)} = \sqrt{2-x}-2\sqrt{2-x} = -\sqrt{2-x}$
$\displaystyle \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}}\:\:\frac{\sqrt{3x+3}+\sqrt{11-x}}{\sqrt{3x+3}+\sqrt{11-x}}$
$\displaystyle \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}\left(\sqrt{3x+3}+\sqrt{11-x}\right)}{4x-8}$
$\displaystyle \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}} = \frac{-\sqrt{2-x}\left(\sqrt{3x+3}+\sqrt{11-x}\right)}{-4(2-x)} = \frac{\sqrt{3x+3}+\sqrt{11-x}}{4\sqrt{2-x}}$
first, note that this limit should be one-sided since the domain of the rational expression is $\displaystyle -1 \le x < 2$
$\displaystyle \lim_{x \to 2^-} \frac{\sqrt{2-x}-\sqrt{8-4x}}{\sqrt{3x+3}-\sqrt{11-x}}$
$\displaystyle \lim_{x \to 2^-} \frac{\sqrt{2-x}-2\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}}$
$\displaystyle \lim_{x \to 2^-} \frac{-\sqrt{2-x}}{\sqrt{3x+3}-\sqrt{11-x}} \cdot \frac{\sqrt{3x+3}+\sqrt{11-x}}{\sqrt{3x+3}+\sqrt{11-x}}$
$\displaystyle \lim_{x \to 2} \frac{-\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{ (3x+3)-(11-x)}$
$\displaystyle \lim_{x \to 2^-} \frac{-\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{4x-8}$
$\displaystyle \lim_{x \to 2} \frac{\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{8-4x}$
$\displaystyle \lim_{x \to 2^-} \frac{\sqrt{2-x} \left[\sqrt{3x+3}+\sqrt{11-x}\right]}{4(2-x)}$
$\displaystyle \lim_{x \to 2^-} \frac{\sqrt{3x+3}+\sqrt{11-x}}{4\sqrt{2-x}}$
as $\displaystyle x \to 2^-$ , the numerator approaches 6 and the denominator approaches 0 ... the limit does not exist.
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