- #1

karush

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ok I presnume the first thing to do is mulitply by conjugate

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- #1

karush

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ok I presnume the first thing to do is mulitply by conjugate

- #2

skeeter

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Yes, rationalizing the denominator would be step 1, resulting in

ok I presnume the first thing to do is mulitply by conjugate

$\displaystyle \lim_{x \to 2} \dfrac{\sqrt{6-x}-2}{2-x} \cdot \lim_{x \to 2} (\sqrt{3-x}+1)$

step 2 would be rationalizing the numerator of the first limit in the product of limits

Step 2 would

- #3

karush

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step 2 would be rationalizing the numerator of the first limit in the product of limits

$\displaystyle \left[ \dfrac{\sqrt{6-x}-2}{2-x}

\cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}

\right]=\frac{\sqrt{-x+3}+1}{\sqrt{-x+6}+2}$

plug in 2

$\dfrac{\sqrt{-2+3}+1}{\sqrt{-2+6}+2}=\dfrac{1}{2}$

hopefully

- #4

skeeter

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$\displaystyle \left[ \dfrac{\sqrt{6-x}-2}{2-x}

\cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}

\right]=\frac{\sqrt{-x+3}+1}{\sqrt{-x+6}+2}$

?

- #5

karush

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?

isn't $\dfrac{1}{2}$ the answer |

- #6

skeeter

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yes, but that quoted equation you posted is not correct

- #7

karush

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as $x \to 2 $ then $=\dfrac{1}{2}$

however I didn't understand the denominator becoming 0

- #8

skeeter

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return to the limit product from post #2

$\displaystyle \lim_{x \to 2} \dfrac{1}{\sqrt{6-x}+2} \cdot \lim_{x \to 2} (\sqrt{3-x}+1)$

$\dfrac{1}{4} \cdot 2 = \dfrac{1}{2}$

- #9

karush

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ok I see the calcular was making the 2nd fraction = to 1

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