If a>1 and ⁿ√a = 1+x, prove that 0 < x < a/n

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if a>1 and ⁿ√a = 1 + x, prove that 0 < x < a/n
Deduce that ⁿ√a →1, n →∞
confused!
 
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If [itex]\sqrt[n]{a}= 1+ x[/itex], then [itex]a= (1+ x)^n[/itex]. Expand using the binomial theorem to get that a= 1+ nx+ positive terms so that 1+ nx< a. Since x< a/n, as n goes to infinity, x goes to 0.
 


hii again :)
thanks also if 0 < a < 1
what is the corresponding result?
 
then n√a would be < 1, so you'd have to write n√a = 1 - x …

so what would happen then? :smile:​