Let $y_1=r^n$, where $r>1$ and $n>1$.
Suppose we decrease $r$ and increase $n$ such that $y_2=(r-\epsilon)^{n+\beta}$. If $\epsilon>\beta$, can we prove that $y_2<y_1$?
Let $y_1=r^n$, where $r>1$ and $n>1$.
Suppose we decrease $r$ and increase $n$ such that $y_2=(r-\epsilon)^{n+\beta}$. If $\epsilon>\beta$, can we prove that $y_2<y_1$?
HINT.-If $0<a<b$ then $a^x<b^x$ for all $x>0$ and $a^x>b^x$ for all $x<0$.
On the other hand, for all $N>0$ you have $x_1$ and $x_2$ such that $$a^{x_1}=b^{x_2}=N$$ You can make your own conclusions (see at the figure).