Question 3.8: If c > 0, then lim c^1/n = 1.

First assume that c > 1, which implies c^{1/n} > 1, so there is a positive d_{n} such that

c^{1/n} = 1+d_{n}.

Using the binomial theorem once again, we obtain

c = (1+d_{n})^{n} > nd_{n},

and therefore

0 < d_{n} < \frac{c}{n} .

From Theorem 3.6 we conclude that d_{n} → 0, and hence

c^{1/n} = 1+d_{n} → 1.

Now assume that 0 < c < 1 and let b = 1 / c. In this case b > 1 so, using the above result, we conclude that b^{1/n} → 1. Now

c^{1/n}=\frac{1}{b^{1/n}} → \frac{1}{1} = 1

by Theorem 3.4.

Finally, if c = 1 then lim c^{1/n} = lim 1 = 1.