Using Sandwich Theorem, how can I evaluate the limit for - $$\lim_{n\to \infty} (a^n+b^n)^{1/n}$$ where $a$ and $b$ are positive quantities and $b$ is greater than $a$.
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$$b < (a^{n} + b^{n})^{1/n} < 2^{1/n}b$$
Seewoo Lee
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Can you please elaborate this? I am getting an idea but not so sure about it. Please help me out. – AdityaS Jan 18 '19 at 18:36
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@user574937 If you can show that the inequality is true, than we get the result from $\lim_{n\to\infty} 2^{1/n} = 2^{\lim_{n\to \infty} 1/n} = 1$ and the sandwitch theorem. To prove the inequality, take $n$th power of both sides and use $a<b$. – Seewoo Lee Jan 18 '19 at 22:24
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That part was understood. I mean, how did you get upon this conclusion? What and how did you thought? – AdityaS Jan 20 '19 at 04:41
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@user574937 If you want some intuition of the proof, since $a<b$, $b^{n}$ will be much bigger than $a^{n}$ when $n$ is very large. So in the sequence $(a^{n} + b^{n})^{1/n}$, $a^{n}$ may not affect the limit. – Seewoo Lee Jan 20 '19 at 04:59