Find limit of $a_n=\sqrt[n]{|(r_n)^n-h^n|}$

Question

Given $g>h>0$, $\lim_{n \to \infty}r_n = g$ and $a_n=\sqrt[n]{|(r_n)^n-h^n|}$. I have to prove that $\lim_{n \to \infty}a_n = g$. I noticed that $a_n \le \sqrt[n]{|r_n^n|+h^n}\le|r_n|$. What to do next?

Answer 1

If $a_n=\sqrt[n]{|(r_n)^n-h^n|}$, then $\dfrac{a_n}{r_n} =\sqrt[n]{|1-(h/r_n)^n|} $.

Since $r_n \to g$, it is enough to show that $\sqrt[n]{|1-(h/r_n)^n|} \to 1$. Since $g > h$, for all large enough $n$ there is a $c < 1$ such that $h/r_n < c$.

So, if you can show that $\sqrt[n]{1-c^n} \to 1$, you are done.

Since $x^n < x$ for $n >1$ and $0 < x < 1$, $x^{1/n} > x$. Therefore $\sqrt[n]{1-c^n} \gt 1-c^n$.

All that is left is to show that $c^n \to 0$.

I will leave that to you.