Find limit of $\sqrt[n]{a^n-b^n}$ as $n\to\infty$, with the initial conditions: $a>b>0$ With the initial conditions: $a>b>0$;
I need to find $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}.$$
I tried to block the equation left and right in order to use the Squeeze (sandwich, two policemen and a drunk, choose your favourite) theorem.
 A: 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. 
A: 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 $$\lim_{n\to\infty}\sqrt[n]{a^n-b^n}=L=a.$$
A: Here is a short solution based on standard inequalities.
Our first inequality is obvious since $b^n>0$ 
$$(1)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\leq a^n.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$ 
Next we note that
$$a^n-b^n = a\cdot a^{n-1}-b\cdot a^{n-1}+ b\cdot a^{n-1}- b\cdot b^{n-1}
   =(a-b)a^{n-1} + b(a^{n-1}-b^{n-1})$$
which together with $a^{n-1}- b^{n-1}\ge0$ leads to 
$$(2)\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad a^n-b^n\geq (a-b)a^{n-1}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $$ 
Combining (1) with (2) and taking the $n$:th-root we get
$$
a \ge (a^n-b^n)^{1/n}\ge(a-b)^{1/n}\cdot a^{(n-1)/n}= a\cdot(a-b)^{1/n}\cdot a^{-1/n}
$$
where the right hand side tends to $a$ as $n\to\infty$, and hence
we reach $$\lim_{n\to\infty} (a^n-b^n)^{1/n}=a.$$
A: Another way: Note that $a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1})$. Since $b < a$, each term in the sum is bounded by $a^{n-1}$ and we have
$$a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} < na^{n-1}$$
Since the sum is at least the first term, we also have
$$a^{n-1} + a^{n-2}b + ... + ab^{n-2} + b^{n-1} > a^{n-1}$$
Combining we have
$$a^{n-1}(a - b) < a^n - b^n < na^{n-1}(a - b)$$
Taking $n$th roots we get
$$a^{1 - {1 \over n}}(a - b)^{1 \over n} < \sqrt[n]{a^n - b^n} <n^{1 \over n} a^{1 - {1 \over n}}(a - b)^{1 \over n} $$
This can be rewritten as
$$a \bigg({a - b \over a}\bigg)^{1 \over n} < \sqrt[n]{a^n - b^n} < a n^{1 \over n} \bigg({a - b \over a}\bigg)^{1 \over n}$$
As $n$ goes to infinity both the left and right sides of the above go to $a$. Thus by the squeeze theorem so does the middle and we are done.
A: Rearrange $(a^n-b^n)^{1/n}$ as $a(1-(b/a)^n)^{1/n}$. Can you do this?
