How Do You Solve a Beginner's Epsilon-Delta Proof for 1/x?

[JPanthon]

Homework Statement

This is my first delt/epsilon proof ever, so please understand if I seem ignorant.

e=epsilon
d = delta

Let f(x) = 1/x for x>0

If e is any positive quantity, find a positive number d, which is such that:

if 0 < |x-2| < d, then |f(x) - 1/2| < e

Homework Equations

I don't really know of any :s

The Attempt at a Solution

|1/x - 1/2| < e
|2/x - 1| < 2e
|x/2 - 1| > 1/2e
|x - 2| > 1/e

and |x-2| < d

Therefore, 1/e < d

Is this sufficient? It says find a positive d, and I've only come up with an inequality with respect to e. Again, this is my first ever d/e proofs, so if I've overlooked some tremendously obvious error, I'm sorry.

 

[LCKurtz]

Homework Statement

This is my first delt/epsilon proof ever, so please understand if I seem ignorant.

e=epsilon
d = delta

Let f(x) = 1/x for x>0

If e is any positive quantity, find a positive number d, which is such that:

if 0 < |x-2| < d, then |f(x) - 1/2| < e

Homework Equations

I don't really know of any :s

The Attempt at a Solution

|1/x - 1/2| < e
|2/x - 1| < 2e
|x/2 - 1| > 1/2e

It's OK to start with an exploratory argument like this, but how did you get from the second step to the third step?

 

[JPanthon]

It's OK to start with an exploratory argument like this, but how did you get from the second step to the third step?

I took the reciprocal of both sides - is that allowed?

 

[LCKurtz]

I took the reciprocal of both sides - is that allowed?

No law against it. Not sure why you want the reciprocal but, given that's what you want to do, do you think the reciprocal of
\frac 2 x -1\hbox{ is }\frac x 2 -1\hbox{?}

 

[JPanthon]

No law against it. Not sure why you want the reciprocal but, given that's what you want to do, do you think the reciprocal of
\frac 2 x -1\hbox{ is }\frac x 2 -1\hbox{?}

Oh, I see my mistake. Thank you.

I don't really know: I took the reciprocal to try to get d and e in the same form, so I may choose d.

Could I have a hint to a more efficient method?

 

[LCKurtz]

Your exploratory start was OK:

\left|\frac 1 x - \frac 1 2\right| &lt; \epsilon

Start by simplifying that left side by combining the fractions. The idea is to see how close x needs to be to 2 to make the inequality work.