Question

How do you prove by definition that the function `f(x)= x^2 sin (1/x)` is continuous at x=0?

Answer 1

Not continuous

Explanation:

This function as is, is not continuous at `x_0=0` because it is not defined there.

`D_f={x` `in` `RR` `:x!=0}` `=` `RR`* `=` `(-oo,0)uu(0,+oo)`

This function would be continuous for example,

`f(x) = {(x^2sin(1/x)", "x!=0),(0" , "x=0):}`

Answer 2

If you re-define the function as Jim suggested, then it would be continuous at `x=0` and this can be proven as show below.

Explanation:

We must show that `lim_{x->0}f(x)=f(0)=0`.

Method 1 (use the `epsilon/delta` definition of a limit):

Let `epsilon>0` be given (this represents an arbitrarily small distance that we'd like the function outputs to be to the limit 0 if `x` is sufficiently close to 0).

Choose `delta=sqrt(epsilon)>0` (this represents the measure of "sufficiently close" and is chosen this way to make the algebra below work out nicely).

Suppose `|x-0| < delta` so that `-sqrt(epsilon) < x < sqrt(epsilon)`. Then `0 leq x^{2} < (sqrt(epsilon))^{2}=epsilon` and `|x^{2}sin(1/x)| < epsilon` (when `x!=0`), since `-1 <= sin(1/x) <= 1` for all `x!=0` (also note that `|f(0)|=0<epsilon`).

But this means that `|f(x)-0|< epsilon` for all `x` such that `|x-0| < delta` and we have shown that `lim_{x->0}f(x)=f(0)=0`, making `f` continuous at `x=0`.

Method 2 (use the Squeeze Theorem ):

The facts that `-1 <= sin(1/x) <=1` for all `x!=0` and `f(0)=0` imply that `-x^{2} <= f(x) <= x^{2}` for all `x`. Since `lim_{x->0}x^{2}=0` and `lim_{x->0}(-x^{2})=0`, it follows that `lim_{x->0}f(x)=0=f(0)` by the Squeeze Theorem.