# Epsilon-Delta Proof problem

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

## 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.

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jbunniii
Homework Helper
Gold Member
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

$$\frac{7}{9x^2 + 15} < \epsilon$$

iff

$$7 < 9x^2 \epsilon + 15 \epsilon$$

iff

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

iff

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

Therefore the following is certainly sufficient:

$$9x^2 > \frac{7}{\epsilon}$$

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.)

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

Cheers.

jbunniii
Homework Helper
Gold Member

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

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

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.

Last edited:
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...

jbunniii