Proving $\frac{x^{2}+1}{x^{3}-9} \rightarrow -\frac{1}{4}$ Using Epsilon-Delta

[italiano91]

The question is to prove \frac{x^{2}+1}{x^{3}-9}\rightarrow-\frac{1}{4} as x\rightarrow1, 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.

 

[SammyS]

The question is to prove \frac{x^{2}+1}{x^{3}-9}\rightarrow-\frac{1}{4} as x\rightarrow1, 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 respectively 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.

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).

 

[italiano91]

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).

Thanks a lot