How can i solve : lim {(cubic root x )-1 ) / (x-1 )} , when x->1 ?

Question

the answer will be 0/0 undefined so we try to manipulate the eqyation
e.g. lim (x^2 * x ) / x the answer will be 0/0 ,
so:
lim (x^2 * x ) / x = lim (x(x-1)) /x =lim x-1=0

Answer 1

You replace the x on the bottom with (cube rt x)^3

Then factor it by difference of 2 cubes:

(cube rt x)^3 - 1^3
= (cube rt x - 1)( cube rt x ^2 + 1 cube rt x + 1^2)

Then the cube rt x - 1 will cancel out leaving

1 over (cube rt x ^2 + 1 cube rt x + 1)

Now you can plug in 1 for x getting 1/3 for the limit

Answer 2

a) The limit is an indeterminate formula 0/0

b) The limit when y tends to 1 : lim(y-1)/(y^3 -1) has the same problem. But here we can divide numerator and denominator by (y-1) and obtain: lim 1/(y^2 + y + 1) and the answer is 1/3

c) Now you manipulate your limit: lim(x^(1/3) -1 )/(x-1)

Change the variable x by y^3 and you will get

lim(y-1)/(y^3-1) and the answer is obtain as explained in b)

Answer 3

lim { ( (cubic root x) -a million ) / x } , while x->0 (x^a million/3-a million)/x The decrease is infinity in view which you have -a million/0. that may no longer an indeterminate case; incredibly that's undefined. As x --> 0 from the superb, y strategies - infinity and as x --> 0 from the left, y strategies + infinity.

Answer 4

lim {x^(1/3) -1 ) / (x-1)} , when x->1
= lim {1/ (x^(2/3)+x^(1/3)+1)} , x->1 use a^3-b^3 = (a-b)(a^2+ab+b^2)
= 1/3
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Or you can use l'Hopital's rule.
lim {x^(1/3) -1 ) / (x-1)} , when x->1
= (1/3)x^(-2/3), x->1
= 1/3