Recurring Styles 4 - Limit of an Expression!

Let $y_n = \sqrt[n^2]{x_n} \longrightarrow \ln(y_n) = \frac{1}{n^2}\ln(x_n)$ .

We need to find $\displaystyle \lim_{n \to \infty} y_n$

$\Longrightarrow \lim_{n \to \infty} \ln(y_n) = \lim_{n \to \infty} \frac{\ln(x_n)}{n^2}$

By Stolz-Cesaro Theorem :

$\lim_{n \to \infty} \frac{\ln(x_n)}{n^2} = \lim_{n \to \infty} \frac{\ln(x_{n+1}) - \ln(x_n)}{(n+1)^2 - n^2}$

$=\lim_{n \to \infty} \frac{\ln(\frac{x_{n+1}}{x_n})}{2n+1}$

Again, by Stolz-Cesaro Theorem :

$= \lim_{n \to \infty} \frac{\ln(\frac{x_{n+2}}{x_{n+1}}) - \ln(\frac{x_{n+1}}{x_n})}{2(n+1)+1 - 2n - 1} = \lim_{n \to \infty} \frac{\ln(\frac{x_n \cdot x_{n+2}}{x_{n+1}^2})}{2} = \frac{\ln(4)}{2} = \ln(2)$

$\Rightarrow \lim_{n \to \infty} \ln(y_n) = \ln(2) \Rightarrow \lim_{n \to \infty} y_n = \boxed{2}$