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

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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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is (x-1) all under the cube root?
No option seem correct
no

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Other answers:

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