Finding a Minimum Value for k in an Epsilon-Delta Proof

[sakodo]

Hey guys here's 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.

 

[jbunniii]

As you said,

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

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

\frac{7}{9x^2 + 15} &lt; \epsilon

iff

7 &lt; 9x^2 \epsilon + 15 \epsilon

iff

9x^2 &gt; \frac{7 - 15\epsilon}{\epsilon}

iff

9 x^2 &gt; \frac{7}{\epsilon} - 15

Therefore the following is certainly sufficient:

9x^2 &gt; \frac{7}{\epsilon}

i.e. if x satisfies this inequality, then it satisfies all the ones above. It should be easy to find a suitable k now. (Note that the problem didn't ask you to find the smallest possible k, just any k that works.)

 

[sakodo]

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.

 

[jbunniii]

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

9x^2 &gt; \frac{7}{\epsilon} - 15

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

9x^2 = \frac{7}{\epsilon} - 15

Let's also set x to the minimum allowed:

x = \frac{k}{\sqrt{\epsilon}}

and substitute this into the boundary case:

\frac{9k^2}{\epsilon} = \frac{7}{\epsilon} - 15

or equivalently

9k^2 = 7 - 15\epsilon

Then we have

k = \frac{1}{3}\sqrt{7 - 15\epsilon}

assuming \epsilon is small enough that we can take the square root.

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

From this we can see that the smallest k that works for ALL \epsilon is

k = \frac{1}{3}\sqrt{7}

which is the same answer as before.

 

[sakodo]

Hey man thanks again. However I just realized 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...

 

[jbunniii]

Hey man thanks again. However I just realized 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.