# How do you explain why f(x) = x^2 if x>=1 and 2x if x<1 is continuous at x = 1?

Nov 30, 2016

It is not continuous at $1$, because the limit at $1$ does not exist.

#### Explanation:

$f \left(x\right) = \left\{\begin{matrix}{x}^{2} & \text{if" & x >=1 \\ 2x & "if} & x < 1\end{matrix}\right.$

$f$ is continuous at $1$ if and only if

${\lim}_{x \rightarrow 1} f \left(x\right) = f \left(1\right)$. $\text{ }$ (existence of both is implied)

We see from the definition of $f$ above that

$f \left(1\right) = {\left(1\right)}^{2} = 1$.

In finding the limit, we'll need to consider the left and right limits separately because the rule changes at $x = 1$.

The limit from the right is

${\lim}_{x \rightarrow {1}^{+}} f \left(x\right) = {\left(1\right)}^{2} = 1$.

The limit from the left is

${\lim}_{x \rightarrow {1}^{-}} f \left(x\right) = 2 \left(1\right) = 2$.

Because the right and left limits are not the same, the limit does not exist.

Therefore the function is not continuous at $1$.