I am not sure what "independent of b". Is it that the value of the limit does not change when b > 0?Show $\displaystyle \lim_{n \rightarrow \infty} b^{\frac{1}{n}}$, for b > 0 is independent of b.

$\displaystyle x = \lim_{n \rightarrow \infty} b^{\frac{1}{n}}$

$\displaystyle \ln (x) = \lim_{n \rightarrow \infty} \ln (b^{\frac{1}{n}})$

$\displaystyle \ln (x) = \lim_{n \rightarrow \infty} \frac{\ln(b)}{n} = 0$

ln(b) is a constant over the the variable. as n approaches infinity the fraction goes to zero. ln(b) is undefined for values <= 0 therefore b > 0 for the limit to be true.

$\displaystyle \ln (x) = 0, x = 1$

$\displaystyle \lim_{n \rightarrow \infty} b^{\frac{1}{n}} = 1$