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Epsilon-Delta Proof problem

  • Thread starter sakodo
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  • #1
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Hey guys heres the problem,

Homework Statement



lim (4x^2+9) / (3x^2 +5) = 4/3
x->infinity

find k, such that x> k/sqrt(epsilon) guarantees abs((4x^2+9) / (3x^2 +5) - 4/3) < epsilon




Homework Equations





The Attempt at a Solution



By removing the absolute sign and making the denominator common, we get

7 / (9x^2 + 15) < epsilon

keep solving we get

x > 1/3 sqrt(7/epsilon - 15)

Here is where I got stuck. How do we find k as a constant? 15 is inside the square root and if we want to make the whole thing as a fraction we would get x > 1/3 sqrt((7-15*epsilon)/epsilon). I can't just get rid of 15 either because that would screw up the inequality and I am meant to find the MINIMUM value of k. Any help would be appreciated.
 

Answers and Replies

  • #2
jbunniii
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As you said,

[tex]\frac{4x^2 + 9}{3x^2 + 5} - \frac{4}{3} = \frac{7}{9x^2 + 15}[/tex]

This is positive for all [itex]x[/itex] so we only need to worry about the upper bound. We have

[tex]\frac{7}{9x^2 + 15} < \epsilon[/tex]

iff

[tex]7 < 9x^2 \epsilon + 15 \epsilon[/tex]

iff

[tex]9x^2 > \frac{7 - 15\epsilon}{\epsilon}[/tex]

iff

[tex]9 x^2 > \frac{7}{\epsilon} - 15[/tex]

Therefore the following is certainly sufficient:

[tex]9x^2 > \frac{7}{\epsilon}[/tex]

i.e. if [itex]x[/itex] satisfies this inequality, then it satisfies all the ones above. It should be easy to find a suitable [itex]k[/itex] now. (Note that the problem didn't ask you to find the smallest possible [itex]k[/itex], just any [itex]k[/itex] that works.)
 
  • #3
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Thanks for the reply jbunnii.

Yeah I get your working. Just a side question, is it then impossible to obtain a minimum value of k?

Cheers.
 
  • #4
jbunniii
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Thanks for the reply jbunnii.

Yeah I get your working. Just a side question, is it then impossible to obtain a minimum value of k?

Cheers.
OK, go back to

[tex]9x^2 > \frac{7}{\epsilon} - 15[/tex]

If this were an equality instead of an inequality, then we would have a "boundary case" where [itex]x[/itex] just barely fails, but any larger [itex]x[/itex] would pass. Let's investigate this boundary case:

[tex]9x^2 = \frac{7}{\epsilon} - 15[/tex]

Let's also set [itex]x[/itex] to the minimum allowed:

[tex]x = \frac{k}{\sqrt{\epsilon}}[/tex]

and substitute this into the boundary case:

[tex]\frac{9k^2}{\epsilon} = \frac{7}{\epsilon} - 15[/tex]

or equivalently

[tex]9k^2 = 7 - 15\epsilon[/tex]

Then we have

[tex]k = \frac{1}{3}\sqrt{7 - 15\epsilon}[/tex]

assuming [itex]\epsilon[/itex] is small enough that we can take the square root.

I think this [itex]k[/itex] is the smallest possible, if [itex]k[/itex] is allowed to depend on [itex]\epsilon[/itex].

From this we can see that the smallest [itex]k[/itex] that works for ALL [itex]\epsilon[/itex] is

[tex]k = \frac{1}{3}\sqrt{7}[/tex]

which is the same answer as before.
 
Last edited:
  • #5
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Hey man thanks again. However I just realised if 9x^2 > 7/epsilon -15, then how can we assume 9x^2 > 7/epsilon? if a > b -c we can't just say a > b can we? grr this is annoying lol...
 
  • #6
jbunniii
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Hey man thanks again. However I just realised if 9x^2 > 7/epsilon -15, then how can we assume 9x^2 > 7/epsilon? if a > b -c we can't just say a > b can we? grr this is annoying lol...
No, you have the logic in reverse.

I established that we need AT LEAST 9x^2 > 7/epsilon - 15. But if we achieve 9x^2 > 7/epsilon, that's even better, right? So I found k to satisfy 9x^2 > 7/epsilon, and therefore it also satisfies 9x^2 > 7/epsilon - 15.
 

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