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I know that $x_n$ is a convergent sequence that converges to the limit $x>0$. The convergence hypothesis implies $$ |x_n-L|<\epsilon, $$ which in this case means $$ |x_n-x|<\epsilon $$ That gives us 2 inequalities which respectively are $$ \begin{align} x_n&<\epsilon+x\\ &\text{and} \\ x_n&>x-\epsilon \end{align} $$ In the first case, if I let $\epsilon=x>0$ then, it gives me $x_n<2x$ and in the second case, if I let $\epsilon=x/2>0$, then $x_n>x/2$. Is that a correct proof and a good way to go about it?

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    $\begingroup$ For the left hand side, let $\varepsilon = x - x/2 = x/2 >0$ and apply the $\varepsilon \delta $ definition of convergence. For the right hand side a similar approach will work. Few people will answer this question, since you have not explained what you tried to solve it on your own. $\endgroup$
    – Thomas
    Oct 20, 2019 at 15:39
  • $\begingroup$ I know, but I was really clueless as to where to start from. What you gave me really helped me! Am I on the right track now? $\endgroup$
    – user716848
    Oct 20, 2019 at 16:06

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The convergence hypothesis implies that

for every $\varepsilon>0$, there exists $N$ such that, for $n\ge N$, $|x_n-x|<\varepsilon$.

You can choose whatever $\varepsilon$ you like. For instance, if we consider $\varepsilon=x/2$, we conclude that there is $N_1$ such that, for $n\ge N_1$, $|x_n-x|<x/2$, that becomes $-x/2<x_n-x<x/2$, in particular $x_n>x/2$.

If we choose $\varepsilon=x$, we find similarly $N_2$ such that, for $n\ge N_2$, $x_n<2x$.

Set $K=\max\{N_1,N_2\}$ and you're done.

In other words, your idea is correct but explained in a quite clumsy, if not incorrect, way

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