Question

Prove that function hasn't lim in `x_0=0` ?

Function:
#f(x)={(2 " when " x=(1,1/2,1/4...)), (1 " when " x=R\(1,1/2,1/4...)):}#

Answer 1

See explanation.

Explanation:

According to Heine's definition of a function limit we have:

`lim_{x->x_0} f(x)=g iff`
`AA{x_n} (lim_{n->+oo}x_n=x_0 => lim_{n->+oo}f(x_n)=g)`

So to show that a function has NO limit at `x_0` we have to find two sequences `{x_n}` and `{bar(x)_n}` such, that

`lim_{n->+oo}x_n=lim_{n->+oo}bar(x)_n=x_0`

and

`lim_{n->+oo}f(x_n)!=lim_{n->+oo}f(bar(x)_n)`

In the given example such sequences can be:

`x_n=1/(2^n)` and `bar(x)_n=1/(3^n)`

Both sequences converge to `x_0=0`, but according to the function's formula we have:

`lim _{n->+oo}f(x_n)=2` (*)
because all elements in `x_n` are in `1,1/2,1/4,...`

and for `bar(x)_n` we have:

`f(bar(x)_1)=f(1)=2`

but for all `n>=2` we have: `f(bar(x)_n)=1`

So for `n->+oo` we have:

`lim_{n->+oo}f(bar(x)_n)=1` (**)

Both sequences coverge to `x_0=0`, but the limits (*) and (**) are NOT equal, so the limit `lim_{x->0}f(x)` does not exist.

QED

The limit definition can be found in Wikipedia at: https://en.wikipedia.org/wiki/Limit_of_a_function

Answer 2

Here is a proof using the negation of the definition of the existence of a limit.

Explanation:

Short version

`f(x)` cannot approach a single number `L` because in any neighborhood of `0`, the function `f` takes on values that differ from each other by `1`.

So no matter what someone proposes for `L`, there are points `x` near `0`, where `f(x)` is at least `1/2` unit away from `L`

Long version

`lim_(xrarr0)f(x)` exists if and only if

there is a number, `L` such the for all `epsilon > 0`, there is a `delta > 0` such that for all `x`, `0 < abs(x) < delta` implies `abs(f(x)-L) < epsilon`

The negation of this is:

`lim_(xrarr0)f(x)` fails to exist if and only if

for every number, `L` there is an `epsilon > 0`, such that for all `delta > 0` there is an `x`, such that `0 < abs(x) < delta` and `abs(f(x)-L) >= epsilon`

Given a number `L`, I will let `epsilon = 1/2` (any smaller `epsilon` will work as well)

Now given a positive `delta`, I must show that there is an `x` with `0 < absx < delta` and `abs(f(x)-L) >= 1/2` (recall that `epsilon = 1/2`)

Given a positive `delta` , eventually `1/2^n < delta` so there is an `x_1` with `f(x_1) = 2`.
There is also an element `x_2 in RR-{1, 1/2, 1/4, . . . }` with `0 < x_2 < delta` and `f(x_2) = 1`

If `L <= (1/2)`, then `abs(f(x_1)-L) >= 1/2`

If `L >= (1/2)`, then `abs(f(x_2)-L) >= 1/2`