Question

Why the function is discontinuous at a=1 given `f(x)= 1-x^2` if x < 1 and `f(x)= 1/x` if `x >= 1`?

Answer 1

The limit as `x` approaches `1` does not exist so it the function is discontinuous at `x=1`.

Explanation:

There are three requirements for a function `f` to be continuous at a `x=a`:

(i) `f(a)` exists
(ii) `lim_(x->a)f(x)` exists
(iii) `f(a)=lim_(x->a)f(x)`

PART 1 We will check if `f(1)` exists. Because `1>=1`, we use `f(x)=1/x`

`f(1)=1/1=1`

`color(blue)(f(a)" exists")`

PART 2 We will check if `lim_(x->a)f(x)` exists. And to do this, we will need to check if the right-hand limit and the left-hand limit are equal.

`lim_(x->a^+)f(x)`
`lim_(x->1^+)1/x=1/1=1`

`lim_(x->a^-)f(x)`
`lim_(x->1^-)(1-x^2)=1-1^2=1-1=0`

Since the right-hand limit and the left-hand limit are not equal, then the limit as `x` approaches `1` does not exist.

The second requirement is not true, therefore the function is discontinuous.