If x, y ∈ R, such that 0 < x < y, then $\underset{n→∞}{\lim}(y^n+x^n)^{\frac{1}{n}}$, is equal to

1 x y e

y

Since $0<x<y∴0<\frac{x}{y}<1$ $∴\underset{n→∞}{\lim}\begin{Bmatrix}\frac{x^n}{y^n}=0\end{Bmatrix}∴\underset{n→∞}{\lim}(y^n+x^n)^{\frac{1}{n}}\underset{n→∞}{\lim}y.e^{\underset{n→∞}{\lim}(\frac{x}{y})^n.\frac{1}{n}}=y.e^0=y$