Epsilon-delta proof

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

I understand the method and get to this stage:

[itex]|x-1|[/itex]|[itex]\frac{x^{2}+5x+5}{4(x^{3}-9)}|[/itex] < ε

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

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 |[itex]\frac{x^{2}+5x+5}{4(x^{3}-9)}|[/itex] using the assumption [itex]|x-1|[/itex] < 1.

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

Using [itex]|x-1|[/itex] < 1 to get -1 < [itex]x-1[/itex] < 1 and then 0 < [itex]x[/itex] < 2, and then taking the numerator [itex]x^{2}+5x+5[/itex] 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 < [itex]x^{2}+5x+5[/itex] < 19 (and hence) 5 < |[itex]x^{2}+5x+5[/itex]| < 19 correct?

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

If so I can get [itex]\frac{4ε}{19}[/itex] and I can finish the proof.

thanks

ps sorry about the messy latex first time using it.
 
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  • #2
SammyS
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The question is to prove [itex]\frac{x^{2}+1}{x^{3}-9}[/itex][itex]\rightarrow[/itex]-[itex]\frac{1}{4}[/itex] as [itex]x[/itex][itex]\rightarrow[/itex]1, using first principles, i.e. using an epsilon-delta proof.

I understand the method and get to this stage:

[itex]|x-1|[/itex][itex]|\frac{x^{2}+5x+5}{4(x^{3}-9)}|[/itex] < ε

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

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 |[itex]\frac{x^{2}+5x+5}{4(x^{3}-9)}|[/itex] using the assumption [itex]|x-1|[/itex] < 1.

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

Using [itex]|x-1|[/itex] < 1 to get -1 < [itex]x-1[/itex] < 1 and then 0 < [itex]x[/itex] < 2, and then taking the numerator [itex]x^{2}+5x+5[/itex] 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 < [itex]x^{2}+5x+5[/itex] < 19 (and hence) 5 < |[itex]x^{2}+5x+5[/itex]| < 19 correct?

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

If so I can get [itex]\frac{4ε}{19}[/itex] 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, [itex]\displaystyle \delta=\min\left(1,\ \frac{4ε}{19}\right)[/itex] is correct, (if that's what you meant).
 
  • #3
The Latex looks fine.

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

Thanks a lot
 

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