how do you solve for the lim(x^3-1)/(x-1) as x approaches 1 algebraically? I know its 3 but im stuck!!?

Answer 1

lim x>1 (x^3-1)/(x-1)
=limx>1 (x-1)(x^2+x+1)/(x-1)
now since x only tends to zero and not
equal to zero we can cancel out thex-1
lim x>1=(1)^2+1+1=3

Answer 2

well dear Kitty its better question ;
lim(x^3-1)/(x-1)
x–> 1
as you know if you put ' +1 ' in this fucntion it gets you 0 but we should try our best not to let it happens . how? i would tell you:

first we need to simplfy (x^3 - 1)
why we do that b czo we need to make anouther ' x - 1 ' that can remove them from numerator & denomiator.
if you dont know who you can make it just x^3 - 1 / x-1
, now lets back
x^3 - 1 = (x-1) ( x^2 +x+1)

lim(x^3-1)/(x-1) = lim (x-1) ( x^2 +x+1) / (x -1) =
x–> 1
{ here you can remove ( x -1), now you can put +1 in the function instead of x }
so we have ;
lim ( x^2 +x+1) = ( 1^2 + 1 +1) = 1+1+1=3
x–> 1
so the answer is ;
lim(x^3-1)/(x-1) = 3
x–> 1

Good Luck & good question.

Answer 3

x^3 -1 = (x - 1)(x^2 + x + 1)
(x^3 -1)/(x-1) = (x^2 + x +1)

lim (x^2 + x + 1) as x tends to 1 is 3 (just substitute 1 in there)

Answer 4

Use L'Hopital's Rule. It says that when the numerator and denominator of a limit both approach zero, the limit is equal to the limit of the derivative of the numerator over the derivative of the denominator. In other words, since 1^3 - 1 = 1 - 1 = 0, you can just take the derivative of the top and the derivative of the bottom, and the limit has the same value.

Answer 5

Factorise the x^3 -1 expressiona and the factor x-1 cancels out.
Substitute 1 for x in the remaining term.

Answer 6

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

(x^3-1) / (x-1) = (x-1) (x^2+x+1) / (x-1)

(x-1) cancels, and we obtain: (x^2+x+1)

lim (x^2+x+1) as x --> 1 is equal to 3

Answer 7

Use La Hospitals since the problem is in indeifinite form by taking the derivative of the numerator and then the denominator.

x^3-1 becomes 3x^2
x -1 becomes 1

3(1)^2 = 3

If this doesn't make sense then check out the section on La Hospital's Rule in your book.

Answer 8

dont know