1. ## Calculus Inequality Conundrum

We're learning about the episilon-delta definition of limits.

Anyway a related question was given by the course. Given $0<|x-1|<2$ what range of values can $|2x-3|$ be?

Well I've got this to $0<|2x-2|<4$ but i really don't know what to do from here, and how to solidly demonstrate the range of values $|2x-3|$ can be.

I managed to show $|2x-3|<5$ in a very long brute force way but I feel like there has to be a better way to do it. Any ideas?

2. ## Re: Calculus Inequality Conundrum

Hey jeremy5561.

Hint: Show that if 0 < |x| < b then 0 <|x-a| < b+a and if 0 < |x| < c then 0 < |dx| < cd.

To get you started consider that 0 < x < b and -b < x < 0 for x > 0 and x < 0 respectively. If I add a this means that:

0 < |x-a| < c which implies 0 < x-a < c and -c < a-x < 0. By shifting the a term out to the boundaries we get:

0 < x < a+c and -c-a < -x < 0. This proves that we have 0 < |x-a| < c+a for a > 0.

Try the same thing for the multiplication case and then combine those two results to show a proof.

3. ## Re: Calculus Inequality Conundrum

Originally Posted by jeremy5561
We're learning about the episilon-delta definition of limits.

Anyway a related question was given by the course. Given $0<|x-1|<2$ what range of values can $|2x-3|$ be?

Well I've got this to $0<|2x-2|<4$ but i really don't know what to do from here, and how to solidly demonstrate the range of values $|2x-3|$ can be.

I managed to show $|2x-3|<5$ in a very long brute force way but I feel like there has to be a better way to do it. Any ideas?
The triangle inequality is useful here... \displaystyle \begin{align*} |a + b| \leq |a| + |b| \end{align*}. So

\displaystyle \begin{align*} |2x - 3| &\leq |2x-2| + |-1| \\ &\leq |2x - 2| + 1 \\ &\leq 4 + 1 \textrm{ since } |2x - 2| < 4 \\ &\leq 5 \end{align*}