Epsilon-Delta Proof problem

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.)

 

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.

 

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.