# Epsilon-delta proof

1. Dec 14, 2011

### italiano91

The question is to prove $\frac{x^{2}+1}{x^{3}-9}$$\rightarrow$-$\frac{1}{4}$ as $x$$\rightarrow$1, using first principles, i.e. using an epsilon-delta proof.

I understand the method and get to this stage:

$|x-1|$|$\frac{x^{2}+5x+5}{4(x^{3}-9)}|$ < ε

which leads to $|x-1|$ < $\frac{ε}{|\frac{x^{2}+5x+5}{4(x^{3}-9)}|}$

Now I need to get the right hand side in terms of ε alone without x, so I know I need to look at the boundaries of |$\frac{x^{2}+5x+5}{4(x^{3}-9)}|$ using the assumption $|x-1|$ < 1.

I want the RHS to be as small as possible so this will be when |$\frac{x^{2}+5x+5}{4(x^{3}-9)}|$ is at its maximum, and this is where I'm unsure if what I've done next is correct.

Using $|x-1|$ < 1 to get -1 < $x-1$ < 1 and then 0 < $x$ < 2, and then taking the numerator $x^{2}+5x+5$ first, is there any problem with putting 0 and 2 respectivley into the formula to get it's min and max values?

ie is 5 < $x^{2}+5x+5$ < 19 (and hence) 5 < |$x^{2}+5x+5$| < 19 correct?

The same applies to $4(x^{3}-9)$, is -36 < $4(x^{3}-9)$ < -4 (and hence) 4 < |$4(x^{3}-9)$| < 36 correct?

If so I can get $\frac{4ε}{19}$ and I can finish the proof.

thanks

ps sorry about the messy latex first time using it.

Last edited: Dec 14, 2011
2. Dec 14, 2011

### SammyS

Staff Emeritus
The Latex looks fine.

I didn't go through all of your derivation. However, the result, $\displaystyle \delta=\min\left(1,\ \frac{4ε}{19}\right)$ is correct, (if that's what you meant).

3. Dec 14, 2011

Thanks a lot