Showing limits using definition of convergence: First year maths. Watch

montyr
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I am stuck on this question...I suck at proofs...and it doesn't help that I have missed the notes on it, also the people I haveasked are clueless too! I would give up but am determined to get it all done tonight! So any hints?

I get that that the limit is zero but just don't get how to prove it.

Any hints/ideas much appreciated.

Cheers, Ryan.
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Kolya
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Well, for it to converge to 0, you need, for all e>0, there existing N(e) such that n>N => |f(x)| < e. So, your task is to try and find the N(e) that will satisfy |f(x)| < e. You can do this by, in your rough working, assuming |f(x)|< e, finding the N(e) that is required, and then going back and showing that if you are given e>0, and set N(e) as your chosen function, then |f(x)|<e follows.
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Gaz031
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As the above post says though replace x by n in his post.
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montyr
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Hmm...I have something similar written down to prove the same but for 1/n but I don't follow it for some reason...I guess it will click eventually but I'm not quite following where the various symbols etc. come from. And the logic beihnd each step, and even how each step follows...I've been trying to get it since I first started the post and before and for some reason cannot ge my head round it, my head feels as though it is going to explode lol. I wanted to actually get some work done tonight...

Replies much appreciated as always!

Thanks, Ryan.
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Gaz031
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The definition is that a_{n} \to a if for each \epsilon &gt; 0 there exists N \in \mathbb N such that n&gt;N \implies |a_{n}-a|&lt;\epsilon.

In the case a_{n}=\frac{1}{n} and showing this tends to :

Let \epsilon &gt; 0.

[Rough working not included in final proof: Want |\frac{1}{n}-0|&lt;\epsilon which happens if and only if n&gt;\frac{1}{\epsilon}].

Take N to be any natural number strictly greater than \frac{1}{\epsilon}. Then n&gt;N \implies |a_{n}-0|=\frac{1}{n}&lt;\frac{1}{N}&lt;\epsi  lon.

So a_{n}\to 0

Now try to see what you need to change for the case a_{n}=\frac{1}{n^2}!
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