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So in solving this question have I messed up the first line.

In further work should the inequality sign on the end of the first line be > instead of <

Thanks

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    Can you not use the algebraic limit theorems or whatever your book calls them? If not then remember that $5/n - 2/n^2 < 5/n$ then use AP and transitivity. – 2017-02-15
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    @lordoftheshadows Isn't it better to check OP's work rather than offer an alternative solution? – 2017-02-15
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    I have two comments. First, instead of writing $1$ divided by $n$ in the way you did before the "so then" part, you should keep it as $\frac{1}{n}$. Same goes for $\frac{1}{n^{2}}$. Also, you've said for all $n \geq N$, $\frac{1}{n^{2}} \leq \frac{1}{n} < \frac{\epsilon}{3}$, and then you said that this implies $\frac{5}{n} - \frac{2}{n^{2}} < \frac{5 \epsilon}{3} - \frac{2 \epsilon}{3}$. Are you sure about this? I need to think about it more, so I'm not saying you're wrong, but are you sure it's right? **EDIT** I've thought about it, and I don't think it's right. – 2017-02-15
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    You can fix the issue I mentioned in my last comment by using the hint provided in the posted answer. – 2017-02-15
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    @user46944 Sorry I didn't make it clear. – 2017-02-15
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    @lordoftheshadows thanks for the reply, I have solved it correctly in terms of limits but was trying to apply archimedean's property - in case of a specific method is required. :) – 2017-02-15
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    Thank you @user46944 excellent advice, was only following how this course book wrote it :) – 2017-02-15

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Hint: $$ \frac{5}{n} - \frac{2}{n^2} \leq \frac{5}{n}. $$

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    Your answer does nothing to communicate to OP whether their work is right or wrong. – 2017-02-15
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    Working with $\frac{2}{n^2}$ just makes the process tedious. – 2017-02-15
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    Thank you so the inequality sign should then be ≤ ? – 2017-02-15
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    Which "$\leq$" are you referring to? For every $\epsilon>0$ there exists an $N$ such that $\frac{5}{n} - \frac{2}{n^2} \leq \frac{5}{n}< \epsilon$ for every $n>N$. – 2017-02-15
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    Great thank you that is what I was asking, but should I have a modulus sign for the LHS of the inequality? – 2017-02-15
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    Okay, glad to be of help. And yes, and you have perfectly used the fact that $|x| < x$. – 2017-02-15