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.

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.
