How Can You Correctly Manipulate Inequalities Involving 1/(n-2) and Epsilon?

[mjjoga]

so I have 1/(n-2). I have that n>max(epsilon+2,1). I need to get 1/(n-2) < epsilon. I know that 1/(n-2)<1/(epsilon+2-2)=1/epsilon. but 1/epsilon is not always less than epsilon. can you see any errors?

 

[darkchild]

Your question is unclear.

 

[Mark44]

so I have 1/(n-2). I have that n>max(epsilon+2,1). I need to get 1/(n-2) < epsilon. I know that 1/(n-2)<1/(epsilon+2-2)=1/epsilon. but 1/epsilon is not always less than epsilon. can you see any errors?

Epsilon is typically a small positive number that is much closer to 0 than it is to 1. Unless epsilon is negative, epsilon + 2 will be larger than 1.

1/(n - 2) < epsilon <==> n - 2 > 1/epsilon, making the reasonable assumptions that n > 2 and epsilon > 0.

If epsilon is small and positive, 1/epsilon is generally larger than epsilon.