Final answer:
To demonstrate that the limit of (|a|^n + |b|^n)^1/n equals max, using the sandwich theorem, assume |a| ≥ |b|. The expression is squeezed between |a| and 2^{1/n}|a|, and as n approaches infinity, both bounds approach |a|, which is max, .
Step-by-step explanation:
To show the limit of (|a|^n + |b|^n)^1/n equals maxa using the sandwich theorem, also known as the squeeze theorem, we should consider both |a| and |b| to be non-negative due to the absolute value. If we assume without loss of generality that |a| ≥ |b|, then |a| is the maximum of the two.
Next, we notice that for all n, |a|^n ≥|b|^n, which implies that |a|^n + |b|^n ≥ |a|^n. Taking the nth root of both sides of this inequality gives us (|a|^n + |b|^n)^1/n ≥ |a|. Similarly, since (|a|^n + |b|^n) ≤ |a|^n + |a|^n = 2|a|^n, taking the nth root gives us (|a|^n + |b|^n)^1/n ≤ (2|a|^n)^1/n = 2^{1/n}|a|.
As n approaches infinity, 2^{1/n} approaches 1, and so both sides of our inequality approach |a|. By the sandwich theorem, the limit of (|a|^n + |b|^n)^1/n as n approaches infinity is |a|, which is the maximum of |a| and |b|. Therefore, the limit of the given expression is indeed max.