$\text{ If }0<\mathrm{h}<\mathrm{q}\text{ then }\underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left({\mathrm{q}}^{\mathrm{n}}+{\mathrm{h}}^{\mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}=$

Given that $0<\mathrm{h}<\mathrm{q}$

The given limit is $\underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left({\mathrm{q}}^{\mathrm{n}}+{\mathrm{h}}^{\mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}$

Since $q$ is bigger take $q^n$ as common 

$\begin{array}{rcl}\underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left({\mathrm{q}}^{\mathrm{n}}+{\mathrm{h}}^{\mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}& =& q\left[\underset{\mathrm{n}\to \mathrm{\infty }}{lim} {\left(1+{\left(\frac{\mathrm{h}}{\mathrm{q}}\right)}^{\mathrm{n}}\right)}^{\frac{1}{\mathrm{n}}}\right]\\ & =& q\left(1\right)\\ & =& q\end{array}$