Limit of x^n+y^n as n -> ∞: max(x, y)

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SUMMARY

The limit of the expression \(\sqrt[n]{x^n+y^n}\) as \(n\) approaches infinity is equal to \(\max(x, y)\) for two positive numbers \(x\) and \(y\). To demonstrate this, assume \(x > y\) and factor \(x^n\) out of the expression, leading to \(\sqrt[n]{x^n(1+(y^n/x^n))}\). By applying the binomial theorem, it can be shown that the limit of the nth root converges to \(x\). The same reasoning applies when \(y > x\).

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show for two positive numbers x,y>0 that

limit for n->infinity : [tex]\sqrt[n]{x^n+y^n}[/tex] = max {x,y}

i don't know how to make a upper boundary(lower boundary is >0 i suppose)
something like assume for instance x bigger than y and than make a boundary with it, but how?
 
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Assume x>y, take x outside the nth root. Use binomial theorem to show the limit of the nth root is 1.
 
x^n = (x^n+y^n) = x^n(1+(y^n)/(x^n))

=>

nth root of (x^n+y^n)=x*nth root of (1+(y^n)/(x^n)), and for x>y, =>x. the same applies for the opposite case.
 

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