Let $0I don't know where to start on this problem. Any guidance would be greatly appreciated. 
 A: I like to use logarithms with problems like this.
If $A = w^{1/k}$ then $\ln A = \frac{1}{k}\ln w$. Here, $w$ is a positive constant.
As $k \to \infty$ we have $\frac{1}{k} \to 0$ and so:
$$\lim_{k \to \infty} \left(\ln A\right) = \lim_{k \to \infty} \left(\frac{1}{k}\ln w\right) = \ln w \left(\lim_{k \to \infty} \frac{1}{k}\right) = \ln w \times 0 = 0$$
If $\ln A \to 0$ as $k \to \infty$ then $A \to \mathrm{e}^0$ as $k \to \infty$. Hence $A \to 1$ as $k \to \infty$.
This is true for all $w > 0$, and hence for $0 < w < 1$.
A: HINT the base of the function(which is exponential) does not change as $k$ changes. Try looking at the graph of $f(x) = a^{\frac{1}{x}}$ and see what happens to $\frac{1}{x}$ as $x$ approaches infinity.
A: Since $0<w<1$ then $w^{1/n}=\frac{1}{1+g_n}$ for some $g_n>0$. By Bernoulli's inequality $$w=\frac{1}{(1+g_n)^n}\leq \frac{1}{(1+ng_n)}<\frac{1}{ng_n}$$
therefore $$0<g_n<1/nw$$ for $n\in \mathbb{N}$.
$$0<1-w^{1/n}=\frac{g_n}{1+g_n}<g_n<1/nw$$
so $$|w^{1/n}-1|<\frac{1}{w}\frac{1}{n}$$
Now as $n\to \infty$ lim $w^{1/n}=1$
