Evaluate the limit, if n>1 is an integer.....

[littlegentleman]

Hi. I am solving this question and it gives me a headache.

Q. Evaluate the limit, if n>1 is an integer,
lim[x-->1] (1-x^(1/2))(1-x^(1/3))(1-x^(1/4)) .....(1-x^(1/n)) / (1-x)^(n-1)

A. I tried to simply the equation by multiplying (1+x^(1/2)) on both deno/nume.
Then, i get lim[x-->1] (1-x)(1-x^(1/3))(1-x^(1/4)) .....(1-x^(1/n)) / (1-x)^(n-1)(1+x^(1/2))
Cancel (1-x) from deno/nume.
then, lim[x-->1] (1-x^(1/3))(1-x^(1/4)) .....(1-x^(1/n)) / (1-x)^(n-2)(1+x^(1/2))
Now, i have a trouble. I can't figure out any way to go further.
Could you give me a hint? Please tell me if my first step is wrong.
Thank you very much.

 

[royhaas]

First consider \(\displaystyle \frac {1-x^{1/(k+1)}}{1-x} , k=1, ... , n-1\).

 

[stapel]

lim[x-->1] (1-x^(1/2))(1-x^(1/3))(1-x^(1/4)) .....(1-x^(1/n)) / (1-x)^(n-1)

If I understand the above correctly, you have n - 1 factors of the form:

. . . . .\(\displaystyle \frac{1\, -\, x^{1/i}}{1\, -\, x}\, =\, \frac{1\, -\, \sqrt{x}}{1\, -\, x}\)

...where i runs from 2 to n - 1.

If n = 2, then, applying the difference of squares formula, you have:

. . . . .\(\displaystyle \frac{1\, -\, \sqrt{x}}{1\, -\, x}\, =\, \frac{1\, -\, \sqrt{x}}{(1\, -\, \sqrt{x})(1\, +\, \sqrt{x}}\, =\, \frac{1}{1\, +\, \sqrt{x}}\)

As x tends toward 1, you get a limit value of 1/2. Can this process be extended, as n gets larger? For n = 3, you can apply the difference of cubes formula and get:

. . . . .\(\displaystyle \frac{(1\, -\, \sqrt{x})(1\, -\, \sqrt[3]{x})}{(1\, -\, x)(1\, -\, x)}\, =\, \left(\frac{1}{1\, +\, \sqrt{x}}\right)\, \left(\frac{1}{1\, +\, \sqrt[3]{x}\, +\, \sqrt[3]{x^2}}\right)\)

As x tends toward 1, you get a limit value of (1/2)(1/3) = 1/6.

At the next step (n = 3), you'd have:

. . . . .\(\displaystyle \left(\frac{1}{1\, +\, \sqrt{x}}\right)\, \left(\frac{1}{1\, +\, \sqrt[3]{x}\, +\, \sqrt[3]{x^2}}\right)\, \left(\frac{1}{(1\, +\, \sqrt[4]{x})(1\, +\, \sqrt[4]{x^2})}\right)\)

...from applying the difference of squares formula twice. The limit here would be (1/2)(1/3)[(1/2)(1/2)] = (1/2)(1/3)(1/4) = 1/24.

I don't know how far this can be taken, or quite how you'd generalize (or prove the generalization of) the above. But the trend looks promising, I think...?

Good luck!

Eliz.

 

[littlegentleman]

Thank you so much!!!