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.

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 $$\lim_{n\to\infty} (a^n-b^n)^{1/n}=a.$$
