Recent content by blunkblot

Presumably the book is using the absolute value to indicate that the solution includes both -inf and +inf. That's how I read it, but I'm happy to know I'm not the only one who finds it bizarre.

Hi, I came across a book which looks at a problem like \lim_{x \to 0}\frac{1}{x} So you approach from 0-, and get -∞, approach from 0+, get ∞ Then it would write the answer as \lim_{x \to 0}\frac{1}{x} = \left| \infty \right| It looks bizarre to me. How do you parse this? Is...

Makes sense, thank you.

The book got it by merging the -5 into the fraction, giving \left| \frac{-4x + 2}{x} \right| Then it maximizes this fraction within the allowed values for x by using 4 for the top x, 1 for the bottom x, and applying absolute value to each term since |-4x + 2| <= |4x| + |2| This gives...

This I understand. If you substitute x with 1, you can get | 3 | + | 5 | = 8 as an upper bound. The solution as linked actually has M = 18. What makes 18 a better upper bound than 8? Or, is the book wrong? Thank you

Apparently I cannot read. I figure this is the wrong thread. Please move or kill as needed :( You need to find M where \left|\frac{x+2}{x}-5\right| \le M, x \in (1, 4) To do this, I put 4 for the top x, 1 for the bottom x, to give the greatest quotient possible, plus |-5|, which finally gives...

I apologize for a misclick double post: Mentor action: double posted thread merged. No harm, no foul.[/color]