Question #158004

Use the definition of limit to prove that the sequence { 1 √ n+1} converges to 0.


1
Expert's answer
2021-01-26T04:15:23-0500

Let ϵ>0Let \space \epsilon >0

We want 1n0.50<ϵ    1n0.5<ϵ    n0.5>1ϵ\mid\frac{1}{n^{0.5}}-0\mid<\epsilon\iff\frac{1}{n^{0.5}}<\epsilon\iff{n^{0.5}}>\frac{1}{\epsilon}

So that n>1ϵ0.5n>\frac{1}{{\epsilon}^{0.5}}

Taking N=1ϵ0.5N = \frac{1}{{\epsilon}^{0.5}} we have n>N     n>1ϵ0.5\implies n>\frac{1}{{\epsilon}^{0.5}} and for which we have

1n0.50<ϵ\mid\frac{1}{n^{0.5}}-0\mid<\epsilon

Meaning limx1n0.5=0\lim\limits_{x\to \infin}\frac{1}{n^{0.5}}=0 by the definition of limit


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