Proof: lim an - bn (n->inf) = M - L

[bavman]

Homework Statement

{a_n} and {b_n} are convergent,

if lim a_n (n->inf) = M and lim b_n (n->inf) = L

Prove that lim a_n -b_n (n->inf) = M - L

Using Epsilon/n proof

Homework Equations

The Attempt at a Solution

Heres my attempt at it:

Given e > 0 we want to find N s.t. for all n>N

|(a_n - b_n) - (M-L)| < e

|(a_n - b_n) - (M-L)| <= |a_n-M|-|b_n-L|

there exist N₁ (element of natural number) s.t. for all n>N, |a_n - M| < e/2

there exist N₂ (element of natural number) s.t. for all n>N, |b_n - L| < e/2

let N = max{N₁, N₂} for all n>N

|(a_n - b_n) - (M-L)| <= |a_n-M|-|b_n-L|<e/2 + e/2 < e

 

[╔(σ_σ)╝]

You applied the triangle inequality incorrectly.

You should have written

$$|a_n - M| + |- ( b_n-L)|$$.
If you have a minus sign between the absolute values then you will run into problems since the difference maybe negative. You want to have an inequality like

0< something < e.


That ensures that you can make "something" as small as you like since it is sandwiched between zero and epsilon.
Apart from that, the logic is sound.