# Limit as x tends to infinity

1. Nov 29, 2008

### sara_87

1. The problem statement, all variables and given/known data

calculate the limit as x tends to infinity of:

$$\sqrt[3]{x}$$ ((x+1)$$^{(2/3)}$$-(x-1)$$^{(2/3)}$$)

2. Relevant equations

3. The attempt at a solution

using the identity: a-b=(a^2-b^2)/(a+b) ; and dividing top and bottom by x,
= lim $$\frac{x^{(1/3)}[(1+1/x)^{(4/3)}-(1-1/x)^{(4/3)}}{(1+1/x)^{(2/3)}+(1-1/x)^{(2/3)}}$$
= infinity
is that right? it seems very wrong.

2. Nov 29, 2008

### Staff: Mentor

The limit is 4/3.
Just to make it simpler to write, let u = (x + 1)^(1/3) and v = (x - 1)^(1/3). Then your expression is
$$x^{1/3}(u^2 - v^2)$$
$$= x^{1/3}(u - v)(u + v)$$
Now multiply by 1 in the form of $$\frac{u^2 + uv + v^2}{u^2 + uv + v^2}$$
This will give you $$x^{1/3}\frac{(u^3 - v^3)(u + v)}{(u^2 + uv + v^2)}$$
Now undo the substitution and take the limit.

3. Nov 29, 2008

### sara_87

thank you

4. Nov 29, 2008

### sara_87

so, that means we have to use the binomial expansion inorder to divide by x^1/3 ??

5. Nov 29, 2008

### Staff: Mentor

No, not at all. You'll have terms with x + 1 and x - 1 to the 1/3 and 2/3 powers. For each of these terms factor as x(1 + 1/x). Depending on the power the original terms are raised to, you'll pull out a factor of x^(1/3) or x^(2/3). In the end, you'll have x to the same power in the numerator as in the denominator, so they cancel.

6. Nov 30, 2008

### sara_87

ok, but then i get:
[2x^(2/3){(1+1/x)^(1/3)+(1-1/x)^(1/3)}]/[x^(2/3){(1+1/x)^(2/3)+(1-1/x)^(2/3)+x^-(1/3)(1+1/x)^(1/3)(1-1/x)^(1/3)
after cancellin the x^(2/3) from top and bottom, the last term in the denominator cancels out as x tends to infinity so isnt the limit 4/2 instead of 4/3 beacuse i get 1+1 at the denominator ??

7. Nov 30, 2008

### Staff: Mentor

My denominator looked like this:
$$x^{2/3}[(1 + 1/x)^{2/3} + (1 + 1/x)^{1/3}(1 - 1/x)^{1/3} + (1 - 1/x)^{2/3}]$$

The first factor of x^(2/3) cancels with the same factor in the numerator. The part in square brackets approaches 3 as x gets large, so I don't understand what you're saying about the last term cancelling.

I'm certain that the limit is 4/3, both from the work I did and verifying the limit with Excel.

8. Nov 30, 2008

### sara_87

oh right, i see. ur absolutely correct and thank you. :)