Calculating the Limit as x Tends to Infinity of a Cube Root Expression

[sara_87]

Homework Statement

calculate the limit as x tends to infinity of:

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

Homework Equations

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.

 

[Mark44]

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.

 

[sara_87]

thank you

 

[sara_87]

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

 

[Mark44]

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

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.

 

[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 isn't the limit 4/2 instead of 4/3 beacuse i get 1+1 at the denominator ??

 

[Mark44]

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.

 

[sara_87]

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