Question

Explain why the function is discontinuous at the given number a = 1 ?

Answer 1

The limit of the first piece as x approaches 1 from the negative side is:

`lim_(x to 1^-) e^x = e ~~ 2.718`

We do need to take a limit for the second piece; we can just evaluate it at 1:

`1^2 = 1`

The function is discontinuous because the two pieces do not have the same y value at 1;

Answer 2

See below

Explanation:

`f(x)` is a piecewise function defined above.

To show that `f(x)` is discontinuous at a given point we can examine the behaviour of the function at that point.

In this case `f(1) = 1^2 = 1` because `f(x) = x^2: x in [1, +oo)`

Now let's consider `f(1-delta)` for some arbitrarily small `delta >0`

In this case `f(1-delta) = e^(1-delta)` because `(1-delta) < 1 forall delta >0`

So as `delta -> 0` then `f(1^-) -> e`

Since `f(1) != f(1^-) -> f(x)` is discontinuous at `x=1`

[NB: `f(1^-)` means the limit of `f(x)` as `x -> 1` from below.]