How to Find the Delta Quantity for an Epsilon and Delta Proof in a 1/x Function?

[zeroheero]

Homework Statement

I am currently having problems with a similar question, and used that post, but I'm finding it hard to solve for x.

the question states. if f(x) = 1/x for every x > 0, there is a positive quantity e (epsilon), find the d(delta) quatity such that

if 0 < l x - 3 l < d, then l 1/x - 1/3 l < e

i simplify l 1/x - 1/3 l < e to l (x - 3)/ 3x l < e, i put the x - 3 in brackets to show that is all over the 3x

then show the relation between d = e, which is if l x - 3 l < d, and l (x - 3)/3x l < e
then l d/3x l = e.

I think I am almost there just need a couple pointers in the right direction i believe.

 

[slider142]

Homework Statement

I am currently having problems with a similar question, and used that post, but I'm finding it hard to solve for x.

the question states. if f(x) = 1/x for every x > 0, there is a positive quantity e (epsilon), find the d(delta) quatity such that

if 0 < l x - 3 l < d, then l 1/x - 1/3 l < e

i simplify l 1/x - 1/3 l < e to l (x - 3)/ 3x l < e, i put the x - 3 in brackets to show that is all over the 3x

then show the relation between d = e, which is if l x - 3 l < d, and l (x - 3)/3x l < e
then l d/3x l = e.

I think I am almost there just need a couple pointers in the right direction i believe.

Use the fact that x is positive to show that l (x - 3)/3x l = |x - 3|/3x < d/3x. Note that d should be defined by an inequality (a continuous range of values for each e), not an equality.
These types of proofs are usually done backwards.
Use a scratch-pad to note that if we want the expression to be bounded by e and the expression is necessarily bounded by d/3x, then we want d/3x < e, which implies d < (some expression involving e).
On the real proof, you then magically say "Let d < (that expression involving e)." and then proceed to show that the expression satisfies the inequality.