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in proving that: [itex]lim_{n\rightarrow\infty} \frac{1}{n}\neq 1[/itex] the following proof was suggested.

Proof:

Suppose [itex]lim_{n\rightarrow\infty}\frac{1}{n} =1[/itex], but [itex]lim_{n\rightarrow\infty}\frac{1}{n} =0[/itex], hence:

For all ε>0

1) There exists mεN such that: [itex]n\geq m\Longrightarrow |\frac{1}{n}|<\frac{\epsilon}{2}[/itex]

2)There exists kεN such that : [itex]n\geq k\Longrightarrow |\frac{1}{n}-1|<\frac{\epsilon}{2}[/itex]

Choose r = max{m,k},then [itex]r\geq m,r\geq k[/itex]

Let ,[itex]n\geq r\Longrightarrow n\geq m\wedge n\geq k[/itex].

Hence : [itex]|\frac{1}{n}|<\frac{\epsilon}{2}[/itex] and [itex]|\frac{1}{n}-1|<\frac{\epsilon}{2}[/itex].

Thus : [itex]|\frac{1}{n}-\frac{1}{n}+1|=1\leq |\frac{1}{n}| + |\frac{1}{n}-1|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon[/itex]

..............or 1<ε..........But since this holds for all ε>0 we put ε=1 and we have 1<1 ,a contradiction .

Therefor [itex]lim_{n\rightarrow\infty}\frac{1}{n}\neq 1[/itex]

Write a formal proof of the above ,thus proving that the above informal proof is wrong

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# Informal vs formal

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