# 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?

Oct 11, 2016

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) $f \left(a\right)$ exists
(ii) ${\lim}_{x \to a} f \left(x\right)$ exists
(iii) $f \left(a\right) = {\lim}_{x \to a} f \left(x\right)$

PART 1 We will check if $f \left(1\right)$ exists. Because $1 \ge 1$, we use $f \left(x\right) = \frac{1}{x}$

$f \left(1\right) = \frac{1}{1} = 1$

$\textcolor{b l u e}{f \left(a\right) \text{ exists}}$

PART 2 We will check if ${\lim}_{x \to a} f \left(x\right)$ exists. And to do this, we will need to check if the right-hand limit and the left-hand limit are equal.

${\lim}_{x \to {a}^{+}} f \left(x\right)$
${\lim}_{x \to {1}^{+}} \frac{1}{x} = \frac{1}{1} = 1$

${\lim}_{x \to {a}^{-}} f \left(x\right)$
${\lim}_{x \to {1}^{-}} \left(1 - {x}^{2}\right) = 1 - {1}^{2} = 1 - 1 = 0$

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.