Use the definition of limit to prove that the sequence { 1 √ n+1} converges to 0.
Let ϵ>0Let \space \epsilon >0Let ϵ>0
We want ∣1n0.5−0∣<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}∣n0.51−0∣<ϵ⟺n0.51<ϵ⟺n0.5>ϵ1
So that n>1ϵ0.5n>\frac{1}{{\epsilon}^{0.5}}n>ϵ0.51
Taking N=1ϵ0.5N = \frac{1}{{\epsilon}^{0.5}}N=ϵ0.51 we have n>N ⟹ n>1ϵ0.5\implies n>\frac{1}{{\epsilon}^{0.5}}⟹n>ϵ0.51 and for which we have
∣1n0.5−0∣<ϵ\mid\frac{1}{n^{0.5}}-0\mid<\epsilon∣n0.51−0∣<ϵ
Meaning limx→∞1n0.5=0\lim\limits_{x\to \infin}\frac{1}{n^{0.5}}=0x→∞limn0.51=0 by the definition of limit
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