# How do you use the definition of continuity to determine weather f is continuous at f(x)= x-4 if x<=0 and x^2+x-4 if x>0?

## The way in which the function $f$ is defined, it seems that its continuity is to be discussed at the pt. $x = 0$.

Aug 19, 2016

$f$ is continuous at $x = 0$, as discussed in the Explanation Section below.

#### Explanation:

The function $f$ is continuous at $x = 0 \iff {\lim}_{x \rightarrow 0} f \left(x\right) = f \left(0\right)$.

As $x \rightarrow 0$ from RHS, i.e., $x \rightarrow 0 + , x > 0 , s o , f \left(x\right) = {x}^{2} + x - 4$, a quadratic poly., known to be cont. on $\mathbb{R}$

$\therefore {\lim}_{x \rightarrow 0 +} f \left(x\right) = {\lim}_{x \rightarrow 0 +} {x}^{2} + x - 4 = - 4. \ldots \left(1\right)$.

As $x \rightarrow 0 - , x < 0 , s o , f \left(x\right) = x - 4 ,$ a linear poly., known to be cont. on $\mathbb{R}$.

$\therefore {\lim}_{x \rightarrow 0 -} f \left(x\right) = {\lim}_{x \rightarrow 0 -} x - 4 = - 4. \ldots \ldots \ldots \ldots . \left(2\right)$.

Also, $f \left(0\right) = {\left[x - 4\right]}_{x = 0} = - 4. \ldots \ldots \ldots \ldots \ldots \left(3\right)$

From $\left(1\right) - \left(3\right) , {\lim}_{x \rightarrow 0} f \left(x\right) = f \left(0\right)$.

Hence, $f$ is continuous at $x = 0$