Tail-call optimization is where you are able to avoid allocating a new stack frame for a function because the calling function will simply return the value that it gets from the called function. The most common use is tail-recursion, where a recursive function written to take advantage of tail-call optimization can use constant stack space.

Scheme is one of the few programming languages that guarantee in the spec that any implementation must provide this optimization, so here are two examples of the factorial function in Scheme:

```
(define (fact x)
(if (= x 0) 1
(* x (fact (- x 1)))))
(define (fact x)
(define (fact-tail x accum)
(if (= x 0) accum
(fact-tail (- x 1) (* x accum))))
(fact-tail x 1))
```

The first function is not tail recursive because when the recursive call is made, the function needs to keep track of the multiplication it needs to do with the result after the call returns. As such, the stack looks as follows:

```
(fact 3)
(* 3 (fact 2))
(* 3 (* 2 (fact 1)))
(* 3 (* 2 (* 1 (fact 0))))
(* 3 (* 2 (* 1 1)))
(* 3 (* 2 1))
(* 3 2)
6
```

In contrast, the stack trace for the tail recursive factorial looks as follows:

```
(fact 3)
(fact-tail 3 1)
(fact-tail 2 3)
(fact-tail 1 6)
(fact-tail 0 6)
6
```

As you can see, we only need to keep track of the same amount of data for every call to fact-tail because we are simply returning the value we get right through to the top. This means that even if I were to call (fact 1000000), I need only the same amount of space as (fact 3). This is not the case with the non-tail-recursive fact, and as such large values may cause a stack overflow.