anonymous
  • anonymous
Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[\lim_{x \rightarrow 2}(\sqrt{x+2}-\sqrt{2x})\div(x^2-2x)\]
anonymous
  • anonymous
@phi
welshfella
  • welshfella
direct substitution gives 0/0 i would apply l'hopitals rule

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anonymous
  • anonymous
I haven't learned about that
anonymous
  • anonymous
this week we just learned about intro to limits
phi
  • phi
one thought is to multiply top and bottom by \( \sqrt{x+2}+ \sqrt{2x} \)
welshfella
  • welshfella
ok so we need to find different way
anonymous
  • anonymous
let me multiply it
phi
  • phi
the top is a difference of squares
anonymous
  • anonymous
|dw:1438036441844:dw|
phi
  • phi
and don't multiply out the bottom you should get \[ \frac{ (x+2) - 2x} { x(x-2)(\sqrt{x+2}+\sqrt{2x})} \]
phi
  • phi
simplify the top
anonymous
  • anonymous
@phi did you multiply the whole equation by the conjugate?
anonymous
  • anonymous
or by what factor?
phi
  • phi
the top was (a-b)(a+b) = a^2 - b^2 where a and b are the square roots the bottom , I just show the multiplication
phi
  • phi
I can't tell what you did , but it looks too complicated to be right.
phi
  • phi
should I go slower?
anonymous
  • anonymous
yes, please. I get where you said that they're difference of squares. But I don't get which factor did you multiply the whole equation to get that result
phi
  • phi
\[ \frac{(\sqrt{x+2}-\sqrt{2x})}{(x^2-2x)} \\\frac{(\sqrt{x+2}-\sqrt{2x})}{x(x-2)} \] (factored the bottom, hopefully something cancels...
phi
  • phi
\[ \frac{(\sqrt{x+2}-\sqrt{2x})}{x(x-2)} \cdot \frac{(\sqrt{x+2}+\sqrt{2x})}{(\sqrt{x+2}+\sqrt{2x})}\]
phi
  • phi
it is easier to see the pattern if we call the first term a and the second b then we have (a-b)(a+b) which we know (right?) is a^2 - b^2
anonymous
  • anonymous
yes
phi
  • phi
or you can FOIL it out. either way the top becomes \[ (\sqrt{x+2})^2 - ( \sqrt{2x})^2 \\ x+2 - 2x \\ 2-x \] or -(x-2)
anonymous
  • anonymous
okay, I got this part.
phi
  • phi
so we now have the limit of \[ \frac{-(x-2)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})}\]
phi
  • phi
as long as x is not *exactly* 2 , we can cancel (x-2) from top and bottom then we can take the limit
anonymous
  • anonymous
-1/8 right?
phi
  • phi
yes
anonymous
  • anonymous
OMG, thank you so much. I really learned everything
phi
  • phi
yw

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