Printable View
Choose δ ∈ (0,1) such that 1 - δ < x < 1 implies 3/M < 2x2 - 3x + 1 < 0; i.e., M/3 > 1/(2x2 - 3x + 1). Notice that 0 < x < 1 also implies 2 < x + 2 < 3. It follows that f(x) = (x+2)/(2x2 - 3x + 1) < M for all 1 - δ < x < 1.
I do not understand the last statement because it seems to say that if 0 < a < b and d < c < 0, then ad < bc. Is this true?
Quote:
Originally Posted by lm1988
Personally, I think that it is unreasonable for you to ask for explanation starting in the middle of someone's else proof.
Why not simply state the actual question and ask for help?
There is no question. I am just asking what the line of reasoning is as in the original post. The only additional information that you need to know is that M < 0 and is in R. This is actually how the proof begins.Quote:
Originally Posted by Plato
This is what I am trying to prove:
http://latex.codecogs.com/gif.latex?...-3x+1}=-\infty
Again, without knowing the reason for that argument, it impossible to comment on the argument itself.Quote:
Originally Posted by lm1988
Here is the best guess I have: if
.
The proof was provided in a textbook, so I am assuming that the author intended for the reader to understand the reason for that argument.Quote:
Originally Posted by PlatoAgain, without knowing the reason for that argument, it impossible to comment on the argument itself.
Here is the best guess I have: if