Question

How do you use the epsilon delta definition to find the limit of `(2n) / (3n+1)` as x approaches `oo`?

Answer 1

You don't.

Explanation:

Find `lim_(xrarroo)(2x)/(3x-1)`

`lim_(xrarroo)(2x)/(3x-1) = lim_(xrarroo)(2)/(3-1/x) = 2/(3-0) = 2/3`

There is no epsilon-delta definition for limits at infinity. A common definition is:

`lim_(xrarroo)f(x) = L` `" "` if and only if

for every positive `epsilon`, there is an `M` such that for every `x>M`, we have `abs(f(x)-L) < epsilon`.

To prove that the limit is `2/3`, we need to show that the definition is satisfied.

In this case, let `M = 2/(3 epsilon)`.

For every `x > 2/(3 epsilon)`, we have `3x+1 > x > 2/(3 epsilon)`.

Therefore, for every `x > 2/(3 epsilon)`, we have `epsilon > 2/(3(3x+1))` and,

since `abs ((2x)/(3x-1)-2/3) = abs ((-2)/(3(3x+1))`, we have

`abs ((2x)/(3x-1)-2/3) < epsilon`