anonymous
  • anonymous
how to do this? lim(x approaches 1) [(x^1/6 -1) / (x^1/3 - 1)] using change of variable..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Using change of variable?
Psymon
  • Psymon
First, change of variable sounds weird. 2nd, can you draw it or put it in the equation editor to be sure what it is?
anonymous
  • anonymous
|dw:1378607466598:dw|

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anonymous
  • anonymous
change of variable is like this: http://www.mymathforum.com/viewtopic.php?f=15&t=32599
Psymon
  • Psymon
Well.....you can actually do a difference of squares believe it or not. You can do difference of squares to the denominator. \[(a-b) = (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})\] \[(x ^{\frac{ 1 }{ 3 }}-1)=(\sqrt{x ^{\frac{ 1 }{ 3 }}}-\sqrt{1})(\sqrt{x ^{\frac{ 1 }{ 3 }}}+\sqrt{1})\] \[[(x ^{\frac{ 1 }{ 3 }})^{\frac{ 1 }{ 2 }}-1)][(x ^{\frac{ 1 }{ 3 }})^{\frac{ 1 }{ 2 }}+1)]\] \[(x ^{\frac{ 1 }{ 6 }}-1)(x ^{\frac{ 1 }{ 6 }}+1)\] Therefore, now that the bottomis factored: \[\lim_{x \rightarrow 1}\frac{ x ^{\frac{ 1 }{ 6 }}-1 }{ (x ^{\frac{ 1 }{ 6 }}-1)(x ^{\frac{ 1 }{ 6 }}+ 1)}\]
anonymous
  • anonymous
i cannot understand what you just wrote, my equation thing is not working..
Psymon
  • Psymon
Oh x_x Forgot, lol. Ill show ya what I did, it was difference of squares.
anonymous
  • anonymous
but i'm supposed to use change of variables???
Psymon
  • Psymon
Alright, I wont do what I was doing then, lol.
Psymon
  • Psymon
|dw:1378608308737:dw|
Psymon
  • Psymon
So that leaves us with 1/(u+2) Now we have to be careful. The limit is as x ->1, but does not indicate what u approaches. So we have to plug in x = 1 into our u-sub. So if u = x^(1/6) - 1 and x goes to 1, then u = (1)^(1/6) - 1, meaning as x goes to 1, u goes to 0. So now you plug in u = 0 into the equation above thats remaining.
anonymous
  • anonymous
thanks! i did a different method but still same|dw:1378609049500:dw|..
Psymon
  • Psymon
You made a mistake with the substituting then. |dw:1378609348750:dw|
anonymous
  • anonymous
aaahhhhh.... i get it.. lol
Psymon
  • Psymon
Yeah, lol. Either way, you had the right idea at least xD
anonymous
  • anonymous
oh ok.. thanks again:))))))

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