can you solve this lim ((x)^2/3-2(x)^1/3+1)/(x-1)^2 when x--->1?

Question

can you solve this lim ((x)^2/3-2(x)^1/3+1)/(x-1)^2 when x--->1

Answer 1

In this you have a limit 0/0, L'hospital says that in this case you have to take the first derivative each part and then evaluate the limit. For the upper part the first derivative is

d {x^(2/3) - 2 x^(1/3) + 1} / dx = (2/3) x^(-1/3) - (2/3) x^(-2/3)

and for the lower part

d {(x-1)^2} / dx = 2 (x - 1)

again, when you want to do the limit, you obtain a 0/0. Just do it again. The next derivative for the upper part is

(-2/9) x^(-4/3) + (4/9) x^(-5/3)

and for the lower part now gives only 2. When you evaluate now, you don't have any strange, just

(-2/9 + 4/9)/2 = 1/9

this is your limit.

Answer 2

Your bracketing is off, and I'm going to assume you want to solve

lim [ x^(2/3) - 2x^(1/3) + 1 ] / [ (x - 1)^2]
x -> 1

One thing to notice about the numerator is that it's a square binomial. Those fractions make it look screwy, but I assure you that it's a perfect square, and it is in fact (x^(1/3) - 1)^2.

lim [ (x^(1/3) - 1)^2 ] / [ (x - 1)^2 ]
x -> 1

Now to perform some more mathematic ingenuity. What you have to do is treat (x - 1) as a difference of cubes. As a reminder, a difference of cubes is factored in this fashion:
(a^3 - b^3) = (a - b) (a^2 + ab + b^2)

lim [ (x^(1/3) - 1)^2 ] / [ (x^(1/3) - 1) (x^(2/3) + x^(1/3) + 1) ]^2
x -> 1

Remember that since all of the denominator is squared, we can square each factor.

lim [ (x^(1/3) - 1)^2 ] / [ (x^(1/3) - 1)^2 (x^(2/3) + x^(1/3) + 1)^2 ]
x -> 1

Now, the numerator beautifully cancels with a factor on the denominator.

lim ( 1 / [x^(2/3) + x^(1/3) + 1]^2
x -> 1

And we can actually plug in x = 1 now:

1 / [1^(2/3) + 1^(1/3) + 1]^2

1 / [1 + 1 + 1]^2

1/(3^2) = 1/9

If you haven't learned L'Hospital's rule yet, this is the ONLY method you can use to solve the limit.

Answer 3

Should be -1/3. Use L'Hopital's rule twice. (Diff top and bottom, compare, repeat) --- you'll get (-6/9)/2.