Find limit of $\sqrt[n]{a^n-b^n}$ as $n\to\infty$, with the initial conditions: $a>b>0$

Rearrange $(a^n-b^n)^{1/n}$ as $a(1-(b/a)^n)^{1/n}$. Can you do this?

METHOD I

Since $a>b>0$, our limit boils down to:

$$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=\lim_{n\to\infty}\sqrt[n]{a^n \left(1-({\frac{b}{a})^{n}}\right)} = \lim_{n\to\infty}\sqrt[n]{a^n}\lim_{n\to\infty}\sqrt[n]{1-(\frac{b}{a})^{n}}=a.$$

METHOD II

We may simply squeeze it:

$$a= \lim_{n\to\infty}\sqrt[n]{(a-b)a^{n-1}}\leq \lim_{n\to\infty}\sqrt[n]{a^n-b^n} \leq \lim_{n\to\infty}\sqrt[n]{a^n}=a$$

The proofs are complete.

Suppose the limit exists and is $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L.$$ Then, $$\begin{align} \log L &=\log\lim_{n\to\infty}\sqrt[n]{a^n-b^n} \\ &=\lim_{n\to\infty}\log\sqrt[n]{a^n-b^n} \\ &=\lim_{n\to\infty}\frac{1}{n}\log(a^n-b^n) \\ &=\lim_{n\to\infty}\frac{\log(a^n-b^n)}{n}. \end{align}$$ If $a>b>1$ (note: $1$, not $0$—I am not sure offhand how to handle the case where $b$ and possibly $a$ are less than 1), $a^n-b^n\to\infty$, so $\log(a^n-b^n)\to\infty$, so the limit is of the indeterminate form $\frac{\infty}{\infty}$ and L'Hôpital's rule applies. $$\begin{align} \lim_{n\to\infty}\frac{\log(a^n-b^n)}{n} &=\lim_{n\to\infty}\frac{\frac{d}{dn}\log(a^n-b^n)}{\frac{d}{dn}n} \\ &=\lim_{n\to\infty}\frac{\frac{1}{a^n-b^n}\cdot\frac{d}{dn}(a^n-b^n)}{1} \\ &=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot(a^n\log a-b^n\log b) \\ &=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot(a^n\log a-b^n\log a+b^n\log a-b^n\log b) \\ &=\lim_{n\to\infty}\frac{1}{a^n-b^n}\cdot((a^n-b^n)\log a+b^n(\log a-\log b)) \\ &=\lim_{n\to\infty}\left(\log a+\frac{b^n}{a^n-b^n}(\log a-\log b)\right) \\ &=\log a+(\log a-\log b)\lim_{n\to\infty}\frac{1}{(\frac{a}{b})^n-1} \end{align}$$ and since $a>b>1$, $\frac{a}{b}>1$, so $(\frac{a}{b})^n\to\infty$ and $\lim_{n\to\infty}\frac{1}{(\frac{a}{b})^n-1}=0$, so $$\log L=\log a$$ and $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L=a.$$