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I need to know the limit of this sequence:

$\lim (a^n +b^n)^{1/n}$ for $a,b>0$

I suppose that this limits will be in function of a and b, but how I can solve it?

José Carlos Santos
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2 Answers2

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Hint:

WLOG $a\ge b$

$a^n<a^n+b^n\le 2a^n$

lab bhattacharjee
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Hint

If $b>a$:

$$\sqrt[n]{a^n+b^n}=b\sqrt[n]{\left(\frac ab\right)^n+1} \Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=b \lim_{n \rightarrow \infty}\sqrt[n]{\left(\frac ab\right)^n+1}$$ $$\Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= b\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {a^n}{b^n}\right)+1}=b$$

If $a>b$:

$$\sqrt[n]{a^n+b^n}=a\sqrt[n]{\left(\frac ba\right)^n+1} \Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=a \lim_{n \rightarrow \infty}\sqrt[n]{\left(\frac ba\right)^n+1}$$ $$\Rightarrow \lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= a\sqrt[n]{\left(\lim_{n \rightarrow \infty}\frac {b^n}{a^n}\right)+1}=a$$ If $a=b$: $$\lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}=a\lim_{n \rightarrow \infty}2^{1/n}=a2^0=a=b$$ In either case we can see that the maximum of $a,b$ happens to be the limit. Thus: $$\lim_{n \rightarrow \infty}\sqrt[n]{a^n+b^n}= \max(a,b)$$

Suhrid Saha
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mfl
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  • No, I don't have to assume that b>a>0, the question asked before assume it. I have to solve it only knowing that $a,b>0$ This limit is $max(a,b)$, but i don't know how to prove it –  Oct 23 '18 at 14:38
  • Well, using @mfl 's idea, out of $a$ and $b$ the one which is bigger will come outside the radical sign and the fraction approaches 0 as $n \rightarrow \infty$ and therefore the limit must be $max(a,b)$ – Suhrid Saha Oct 23 '18 at 17:19