anonymous
  • anonymous
given that f(x) =(∛x-1)/(x-1) then its limit as x approaches 1 is equal to? a.0 b.1 c.-1 d.does not exist
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
is (x-1) all under the cube root?
anonymous
  • anonymous
No option seem correct
anonymous
  • anonymous
no

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More answers

anonymous
  • anonymous
x-1 not under the cube root
myininaya
  • myininaya
\[\text{ let } u=x^\frac{1}{3} ; x->1 => u->(1)^\frac{1}{3}=1\] so we have \[\lim_{u \rightarrow 1}\frac{u-1}{u^3-1}=\lim_{u \rightarrow 1}\frac{u-1}{(u-1)(u^2+u+1)}\]
myininaya
  • myininaya
that should help out
nenadmatematika
  • nenadmatematika
solution is 1/3 if you wrote it correctly
anonymous
  • anonymous
so if theres no answer i put E.
nenadmatematika
  • nenadmatematika
I suggest you type it again more carefuly and post again...so I would be sure
anonymous
  • anonymous
It's 0, look at myininaya's solution... \[\frac{1-1}{1^2+1+1}=\frac{0}{3}=0\]
myininaya
  • myininaya
hey rick that top part cancels with (x-1) though
myininaya
  • myininaya
i mean u-1
anonymous
  • anonymous
ah right, my mistake
myininaya
  • myininaya
so you do get 1/3
nenadmatematika
  • nenadmatematika
Nenad is the champ :D
nenadmatematika
  • nenadmatematika
hahaha...kidding
anonymous
  • anonymous
lol
anonymous
  • anonymous
this one has a sqr (∛x-1) the other 1 hast no cube root (x-1) and i feel nenad is the champ XD + myiniaya XD
myininaya
  • myininaya
\[\lim_{x \rightarrow 1}\frac{\sqrt[3]{x-1}}{x-1}\] are you saying this is the problem?
anonymous
  • anonymous
that's what i asked him earlier..he said no
anonymous
  • anonymous
YES THATS the one thanks
anonymous
  • anonymous
sorry rick im confused
anonymous
  • anonymous
so what is my answer d or e? e for error
myininaya
  • myininaya
\[\lim_{x \rightarrow 1}\frac{\sqrt[3]{x-1} \cdot (x-1)^3}{(x-1) \cdot (x-1)^3}\] \[\lim_{x \rightarrow 1}\frac{(x-1)}{(x-1)(x-1)^3}=\lim_{x \rightarrow 1}\frac{1}{(x-1)^4}\] I get the limit does not exist
nenadmatematika
  • nenadmatematika
does not exist
anonymous
  • anonymous
or \[\infty\]
anonymous
  • anonymous
thanks rick
anonymous
  • anonymous
d is wrong to i got the result of my question i got 16/20 =90 ^^ thanks to all of u

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