Question

`lim_(x->0)sin(1/x)/(sin(1/x))` ?

Find the limit `lim_(x->0)sin(1/x)/(sin(1/x))`

How would you approach this? Is it `1` or it doesn't exist?

Answer 1

`lim_(x rarr 0) \ sin(1/x)/(sin(1/x)) = 1`

Explanation:

we seek:

`L = lim_(x rarr 0) \ sin(1/x)/(sin(1/x))`

When we evaluate a limit we look at the behaviour of the function "near" the point, not necessarily the behaviour of the function "at" the point in question, thus as `x rarr 0`, at no point do we need to consider what happens at `x=0`, Thus we get the trivial result:

`L = lim_(x rarr 0) \ sin(1/x)/(sin(1/x))`

`\ \ = lim_(x rarr 0) \ 1`

`\ \ = 1`

For clarity a graph of the function to visualise the behaviour around `x=0`
graph{sin(1/x)/sin(1/x) [-10, 10, -5, 5]}

It should be made clear that the function `y=sin(1/x)/sin(1/x)` is undefined at `x=0`

Answer 2

Please see below.

Explanation:

The definitions of limit of a function I use are equivalent to:

`lim_(xrarra)f(x) = L` if and only of For every positive `epsilon`, there is a positive `delta` such that for every `x`, if `0 < abs(x-a) < delta` then `abs(f(x) - L) < epsilon`

Because of the meaning of "`abs(f(x) - L) < epsilon`", this requires that for all `x` with `0 < abs(x-a) < delta`, `f(x)` is defined.

That is, for the required `delta`, all of `(a-delta,a+delta)` except possibly `a`, lies in the domain of `f`.

All of this the gets us:

`lim_(xrarra)f(x)` exists only if `f` is defined in some open interval containing `a`, except perhaps at `a`.

(`f` must be defined in some deleted open neighborhood of `a`)

Therefore, `lim_(xrarr0)sin(1/x)/sin(1/x)` does not exist.

A nearly trivial example

`f(x) = 1` for `x` an irrational real (undefined for rationals)

`lim_(xrarr0) f(x)` does not exist.