Proving sequence tends to limit

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  • #1
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Homework Statement


Prove that $$\frac{x_n^2 - e}{x_n} \rightarrow 1-e$$ as ##n \rightarrow \infty##, provided ##x_n \rightarrow 1## as n ##\rightarrow \infty##.

The Attempt at a Solution


The above holds if ##\,\forall\, \epsilon > 0 \,\exists \, N\, \in\, \mathbb{N}## such that if n ##\geq N## then ##|\frac{x_n^2 - e}{x_n} - (1-e)| < \epsilon##. Work with the expression in the mod signs and rewrite as: $$|x_n - \frac{e}{x_n} - 1 +e| \leq |x_n - 1| + |e-\frac{e}{x_n}|$$using the ##\Delta## inequality. The first term is ##< \epsilon## and the additional term on the RHS is positive, so can I just stop here?

If not, I proceeded using the fact that ##|x_n-1| < \epsilon## and established that ##|-\frac{1}{x_n}| < \frac{1}{1-\epsilon}\,\,(1)## This gives an upper bound entiirely in terms of ##\epsilon##. Since the statment should hold for all ##\epsilon##,then I would usually finsih here. However, I notice that if ##\epsilon =1## then I end up with nonsense. Is there something else I can try?

Many thanks.
 

Answers and Replies

  • #2
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The first term is <ϵ and the additional term on the RHS is positive, so can I just stop here?
You cannot prove that something is smaller than ϵ with the statement "it is smaller than ϵ+(something positive)".

Do you know some basic laws of limits? Like the limit of a_n+b_n, if both a_n and b_n converge? You can prove the statement with those laws only, you don't need any ϵ.
 
  • #3
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You cannot prove that something is smaller than ϵ with the statement "it is smaller than ϵ+(something positive)".

I have seen proofs in my book where they have LHS < 2ε and conclude even though 2ε is greater than ε since ε is positive. The reasoning was that the statement holds for all epsilon, so it satisfies the requirement for the LHS to converge. I was trying to apply the same reasoning here: since above we have LHS < ε + ε, then in my case, the second terms is also > 0 so it is quite similar, no?

Do you know some basic laws of limits? Like the limit of a_n+b_n, if both a_n and b_n converge? You can prove the statement with those laws only, you don't need any ϵ.

Yes, but I should have made it clear in the problem statement that this is to be done via ε-N.
 
  • #4
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I have seen proofs in my book where they have LHS < 2ε and conclude even though 2ε is greater than ε since ε is positive. The reasoning was that the statement holds for all epsilon, so it satisfies the requirement for the LHS to converge. I was trying to apply the same reasoning here: since above we have LHS < ε + ε, then in my case, the second terms is also > 0 so it is quite similar, no?
It works if you include that factor of 2 and find an upper estimate for the second factor, right.

You can restrict the analysis to ϵ<a for any number a you like - in particular, it is not necessary to consider ϵ=1.
 
  • #5
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It works if you include that factor of 2 and find an upper estimate for the second factor, right.

You can restrict the analysis to ϵ<a for any number a you like - in particular, it is not necessary to consider ϵ=1.

Since the statement is to hold for every ε > 0, though, when I take ε to be 1, the upper bound for |1/xn| is undefined. So the bound I have is not true for all epsilon.
 
  • #6
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That is not the direction the logic works.
"For every ε, there is an N such that |whatever(n)|<ε for n>N"
If that is true for ε=0.1 (with N=124, for example), it means that |whatever(n)|<0.1 for n>124.
This implies |whatever(n)|<1 for n>124, so ε=1 is fine as well and you don't have to check it.

Therefore, you can restrict the analysis to small ε if you like.
 
  • #7
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Thanks mfb, so just going back to my workings in the OP, I have established an upper bound in terms of ##\epsilon## and so the LHS will be smaller than some small ##\epsilon##. However, since i have e in the second term: (my upper bound is) $$LHS < \epsilon + e \frac{2-\epsilon}{1-\epsilon},$$ I can't make the LHS get any smaller than ##\approx e##.
 
  • #8
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I don't know where that expression comes from.

##|x_n-x|<a## => ##|\frac{1}{x_n}-\frac{1}{x}|<\frac{2a}{x^2}## for sufficiently small a and xn>0 ,x>0 could be a useful approximation.
 
  • #9
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I don't know where that expression comes from.

I got $$|\frac{x_n^2 - e}{x_n} - (1-e)| \leq |x_n - 1| + |e - \frac{e}{x_n}|$$ as in the OP by triangle inequality. I have an upper bound for the first term and to get an upper bound for the second I applied the triangle inequality again: $$|e - \frac{e}{x_n}| = e|1-\frac{1}{x_n}| \leq e[1 + |-\frac{1}{x_n}|]\,\,(1)$$ Using the fact that ##-\epsilon < x_n - 1 < \epsilon## I have that ##|-\frac{1}{x_n}| < \frac{1}{1-\epsilon}##. So (1) is < e + (e/1-ε)

Then $$|\frac{x_n^2 - e}{x_n} - (1-e)| \leq \epsilon + e\left(1 + \frac{1}{1-\epsilon}\right) = \epsilon + e\left(\frac{2 - \epsilon}{1-\epsilon}\right)$$
 
  • #10
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Well, that method does not work, as you can see from the result. You need some better way to re-write 1/x_n.
 
  • #11
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Well, that method does not work, as you can see from the result.

It doesn't work because we have a lower bound for the RHS (that being e which is not equal to 0)?
You need some better way to re-write 1/x_n.

Instead of using the triangle inequality again ,I rewrote $$|1-\frac{1}{x_n}| = \left|\frac{x_n - 1}{x_n}\right| = \frac{|x_n - 1|}{|x_n|} < \frac{\epsilon}{x_n}$$ For ##n \geq N##, ##x_n \rightarrow 1##, so this is less than ##\epsilon##. Is it better?

EDIT: I removed the mod sign for some reason. I am not sure how else to express 1/x_n. I got the bound $$\left|1-\frac{1}{x_n}\right| < \frac{\epsilon}{1+\epsilon}$$ so in the end, I have a final bound of $$LHS < \epsilon + \frac{e\epsilon}{1+\epsilon}$$ but this still limits the size of the right hand side.
 
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  • #12
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It doesn't work because we have a lower bound for the RHS (that being e which is not equal to 0)?
Maybe it is clearer with an example:
If you want to show that 1<3 and you show 1<5 this inequality is right, but it does not help. Showing 1<2 would help, however (assuming you know 2<3).

Instead of using the triangle inequality again ,I rewrote $$|1-\frac{1}{x_n}| = \left|\frac{x_n - 1}{x_n}\right| = \frac{|x_n - 1|}{|x_n|} < \frac{\epsilon}{x_n}$$ For ##n \geq N##, ##x_n \rightarrow 1##, so this is less than ##\epsilon##. Is it better?
x_n->1 is not sufficient to have ##\frac{\epsilon}{x_n}<\epsilon##.

The last formula looks better, but I think it uses the wrong conclusion from above. It should work with 1-ϵ in the denominator.

You still need some variation to get the whole thing to be <ϵ, or modify the start of the proof such that <ϵ*whatever is sufficient.
 

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