Question

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

Answer 1

is (x-1) all under the cube root?

Answer 2

No option seem correct

Answer 3

no

Answer 4

x-1 not under the cube root

Answer 5

\[\text{ let } u=x^\frac{1}{3} ; x->1 => u->(1)^\frac{1}{3}=1\]
so we have
\[\lim_{u \rightarrow 1}\frac{u-1}{u^3-1}=\lim_{u \rightarrow 1}\frac{u-1}{(u-1)(u^2+u+1)}\]

Answer 6

that should help out

Answer 7

solution is 1/3 if you wrote it correctly

Answer 8

so if theres no answer i put E.

Answer 9

I suggest you type it again more carefuly and post again...so I would be sure

Answer 10

It's 0, look at myininaya's solution...
\[\frac{1-1}{1^2+1+1}=\frac{0}{3}=0\]

Answer 11

hey rick that top part cancels with (x-1) though

Answer 12

i mean u-1

Answer 13

ah right, my mistake

Answer 14

so you do get 1/3

Answer 15

Nenad is the champ :D

Answer 16

hahaha...kidding

Answer 17

lol

Answer 18

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

Answer 19

\[\lim_{x \rightarrow 1}\frac{\sqrt[3]{x-1}}{x-1}\]
are you saying this is the problem?

Answer 20

that's what i asked him earlier..he said no

Answer 21

YES THATS the one thanks

Answer 22

sorry rick im confused

Answer 23

so what is my answer d or e? e for error

Answer 24

\[\lim_{x \rightarrow 1}\frac{\sqrt[3]{x-1} \cdot (x-1)^3}{(x-1) \cdot (x-1)^3}\]
\[\lim_{x \rightarrow 1}\frac{(x-1)}{(x-1)(x-1)^3}=\lim_{x \rightarrow 1}\frac{1}{(x-1)^4}\]
I get the limit does not exist

Answer 25

does not exist

Answer 26

or \[\infty\]

Answer 27

thanks rick

Answer 28

d is wrong to i got the result of my question i got 16/20 =90 ^^ thanks to all of u