How do you prove the statement lim as x approaches 9 for `root4(9-x) = 0` using the epsilon and delta definition?

The `delta"-"epsilon` definition of a limit states that:

the limit of a function `f(x)`, as `x` approaches some value `c`, is `L` if, for every possible `epsilon>0`, we can find a `delta>0` that depends on `epsilon`, such that `abs(f(x)-L) < epsilon` whenever `abs(x-c) < delta`.

It's like a game. Player 1 picks an `epsilon>0`, and Player 2 is trying to find a `delta>0` such that every `x` within `+-delta` of `c` (that is, `color(navy)(x in (c-delta, c+delta))`) is guaranteed to get mapped to an `f(x)` within `+-epsilon` of `L` (that is, `color(green)(f(x) in (L-epsilon, L+epsilon))`). Player 1 keeps picking smaller and smaller `epsilon`, and Player 2 keeps having to find smaller and smaller `delta`.

If we can prove that, for every `epsilon` Player 1 picks, Player 2 can find a suitable `delta`, then we've proven the limit. In other words:

If

`AA" "epsilon > 0" "EE" "delta>0"`

such that

`abs(f(x)-L) < epsilon` when `abs(x-c) < delta`

then

`lim_(x->c)f(x)=L`.

Okay, so that's a lot of mumble jumble. Let's put it to use.

We need to find a `delta` that depends on `epsilon`. That means you can think of `delta` as a function of `epsilon`. We also want to have `'x " is within "delta" of "c'` imply `'f(x)" is within " epsilon " of "L'`, or in math lingo:

`abs(x-c) < delta => abs(f(x)-L) < epsilon`

So we start with `abs(x-c) < delta` and `color(blue)("try to get it to look like " abs(f(x)-L) < epsilon)`.

In this example, `c=9`, `L=0`, and `f(x)=root(4)(9-x)`.

`abs(x-c) < delta => abs(x-9) < delta`
`color(white)(abs(x-c) < delta) => abs(9-x) < delta`
`color(white)(abs(x-c) < delta) => root(4)abs(9-x) < root(4)delta`
`color(white)(abs(x-c) < delta) => abs(root(4)(9-x)) < root(4)delta`
`color(white)(abs(x-c) < delta) => abs(root(4)(9-x)-0) < root(4)delta`
`color(white)(abs(x-c) < delta) => abs(f(x)-0) < root(4)delta`

Hey—looks like we may have found our connection between `delta` and `epsilon`! If we let `root(4)delta = epsilon`, then we have

`color(white)(abs(x-c) < delta) => abs(f(x)-0) < epsilon`

and so we've shown that `abs(x-9) < delta => abs(f(x)-0) < epsilon`.

The last thing to do is to solve `root(4)delta = epsilon` for `delta`:

`"   "root(4)delta=epsilon`
`=> delta = epsilon^4`

All the work we've done here simply means that no matter how small an `epsilon` Player 1 may pick, Player 2 can always just choose their `delta` to be `epsilon^4`, and they'll win every time. That is, as long as `x` is within `epsilon^4` of `9`, `root(4)(9-x)` will be within `epsilon` of `0`. Thus,

`lim_(x->9)root(4)(9-x)=0`

has been proven.

`QED.`