Question

How do you use the formal definition of a limit to prove `lim (x/(x-3)) =1` as x approaches infinity?

Answer 1

`lim_(x->oo)x/(x-3) = 1`

Explanation:

If we look at the graph of `y=x/(x-3)` we can see that it is clear that the limit exists, and is approximately `1`

graph{x/(x-3) [-30, 30, -2, 2]}

Now, As `x->oo` then `1/x->0`

So, it would be better if we could replace `x` with `1/x`

`lim_(x->oo)x/(x-3) = lim_(x->oo)x/(x-3)*(1/x)/(1/x)`

`:. lim_(x->oo)x/(x-3) = lim_(x->oo)(1/x x)/(1/x(x-3))`

`:. lim_(x->oo)x/(x-3) = lim_(x->oo)(1)/(1-3/x)`

And, using `1/x->0` as `x->oo` we have;

`:. lim_(x->oo)x/(x-3) = (1)/(1-0) = 1` QED

Which is completely consistent with the above graph.

Answer 2

Please see below.

Explanation:

I take the formal defintion to be:

`lim_(xrarroo)f(x) = L`

if and only if

for every positive `epsilon`, there is an `M` such that for all `x`, if `x > M`, then `abs(f(x)-L) < epsilon`

Preliminary investigation To show that `lim_(xrarroo)x/(x-3) =1`, we need to make `abs(x/(x-3) - 1)` small, by making `x` large.

`abs(x/(x-3) - 1) = abs (x/(x-3) - (x-3)/(x-3)) = abs(3/(x-3)) = 3/abs(x-3)`

If we make sure that `M > 3`, the we will have `abs(x/(x-3) - 1) = 3/(x-3)`

So we need to make `3/(x-3) < epsilon` and we will make sure that `x-3` is positive, so we'll be OK if we make `3/epsilon < x-3` and `3/epsilon+3 < x`.

So our `M` will be `3/epsilon+3` (or a number greater than that).

Claim: `lim_(xrarroo)x/(x-3) =1`

Proof:

Given `epsilon > 0`, let `M = 3/epsilon+3`

(Note that `M > 3` since `3/epsilon` is positive.)

For any `x > M`, we have `x -3 > 3/epsilon`, so `x-3 > 0` and

`epsilon (x-3) > 3` and, finally `epsilon > 3/(x-3)`

Observe, now, that for `x > M`,

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

Therefore, by the definition of limit at infinity,

`lim_(xrarroo)x/(x-3) =1`.