Question

By the definition of continuity, how do you show that `xsin(1/x)` is continuous at x=0?

Answer 1

It is not continuous at `0`.

Explanation:

`f` is continuous at `a` if and only if `lim_(xrarra)f(x) = f(a)`

If `f(a)` does not exist, then `f` is not continuous at `a`.

Since `0sin(1/0)` does not exists, `xsin(1/x)` is not continuous at `0`.

Answer 2

The function as defined is not, but a slight modification is if we define a modified function by:

`f(x) = { ( xsin(1/x), x != 0), (0,x=0):}`

Explanation:

We will need the definition of continuity which is that:

`f(x)` is continuous at `x=a iff lim_(x rarr a)f(x)=f(a)`

So, in order to prove that the function defined by:

`f(x) = xsin (1/x)`

Is continuous at `x=0` we must show that

`lim_(x rarr 0)xsin(1/x) = f(0)`

This leads is to an immediate problem as `f(0)` is clearly undefined.

This is, however, not the end of the problem. If we look at the graph of the function, it certainly looks as though the function `f(x)` is continuous:

graph{xsin(1/x) [-0.2034, 0.2068, -0.1015, 0.1036]}

Firstly, Let us try and establish if the above limit exists

We can very easily show the limit exists and find its value:

Method 1:

Let `z = 1/x` then as `x rarr 0 => z rarr oo`

So then, the limit can be written:

`lim_(x rarr 0)xsin(1/x) = lim_(z rarr oo) (1/z)sinz`
`" " = lim_(z rarr oo) (sinz)/z`
`" " = 0`

As `|sin(z)| le 1` and `1/z rarr 0` as `z rarr oo`

Method 2
We can also the squeeze (or sandwich) theorem. We know `sinx` oscillates between `+-1`, that is:,

`-1 le sintheta le 1`

Let `theta = 1/x` then

`-1 le sin(1/x) le 1`

Multiply both sides by `x`, then depending upon the sign of `x` we have:

`x >0 => -x <= xsin(1/x) <= x`
`x<0 => -x >= xsin(1/x) >= x`

And since:

`lim_(x rarr 0^+) x = lim_(x rarr 0^+) -x = 0`
`lim_(x rarr 0^-) x = lim_(x rarr 0^-) -x = 0`

We can use the squeeze theorem to say

`lim_(x rarr 0) xsin(1/x) = 0`

So how does this help us? We have shown the limit exists, but we still have the issue with `f(0)` being undefined. We can easily deal with this by modifying the function to explicitly define `f(0)=0` as follows:

`f(x) = { ( xsin(1/x), x != 0), (0,x=0):}`