-1.6.Limit with radicals

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  • #1
karush
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find limit $\displaystyle \lim_{x \to 2}\dfrac{\sqrt{6-x}{-2}}{\sqrt{3-x}-1}$
ok I presnume the first thing to do is mulitply by conjugate
 

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  • #2
skeeter
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find limit $\displaystyle \lim_{x \to 2}\dfrac{\sqrt{6-x}{-2}}{\sqrt{3-x}-1}$
ok I presnume the first thing to do is mulitply by conjugate
Yes, rationalizing the denominator would be step 1, resulting in

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

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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$\displaystyle \left[ \dfrac{\sqrt{6-x}-2}{2-x}
\cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2}
\right]=\frac{\sqrt{-x+3}+1}{\sqrt{-x+6}+2}$

?
 
  • #6
skeeter
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yes, but that quoted equation you posted is not correct
 
  • #7
karush
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$\:\left[\frac{\sqrt{6-x}-2}{2-x}\cdot \frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}\right]:\quad \frac{\sqrt{6-x}-2}{2-x}$

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

however I didn't understand the denominator becoming 0
 
  • #8
skeeter
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$\dfrac{\sqrt{6-x}-2}{2-x} \cdot \dfrac{\sqrt{6-x}+2}{\sqrt{6-x}+2} = \dfrac{(6-x)-4}{(2-x)(\sqrt{6-x}+2)} = \dfrac{\cancel{2-x}}{\cancel{(2-x)}(\sqrt{6-x}+2)} = \dfrac{1}{\sqrt{6-x}+2}$

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
...
A project I am developing ,,, its a lot more than I thot

https://dl.orangedox.com/EAwBaH3HAWsQbCDhxF
stn00.png

Surf the Nations back Entry
 
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