# If g(x)=x if x<0, x^2 if 0<= x <=1, x^3 if x>1, how do you show that g is continuous on (all real #'s)?

Mar 1, 2017

The function $g \left(x\right)$ is defined as follows $\forall x \in \mathbb{R}$:

$g \left(x\right) = \left\{\begin{matrix}x & x < 0 \\ {x}^{2} & 0 \le x \le 1 \\ {x}^{3} & x > 1\end{matrix}\right.$

Which we can graph as follows:

The individual functions $y = x , {x}^{2}$ and ${x}^{3}$ (being polynomials) are all continuous. In order to prove that $g \left(x\right)$ is continuous we need to establish continuity at the connection points between the various curve segments, ie at $x = 0$ and $x = 1$.

When $x = 0$

Consider the left-hand limit:

${\lim}_{x \rightarrow {0}^{-}} g \left(x\right) = {\lim}_{x \rightarrow 0} x \setminus \setminus = 0$

And the right-hand limit

${\lim}_{x \rightarrow {0}^{+}} g \left(x\right) = {\lim}_{x \rightarrow 0} {x}^{2} = 0$

And the value of the function:

$g \left(0\right) = 0$

So $g \left(x\right)$ is continuous at $x = 0$

When $x = 1$

Consider the left-hand limit:

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

And the right-hand limit

${\lim}_{x \rightarrow {1}^{+}} g \left(x\right) = {\lim}_{x \rightarrow 1} {x}^{3} = 1$

And the value of the function:

$g \left(1\right) = 1$

So $g \left(x\right)$ is continuous at $x = 1$

Hence $g \left(x\right)$ is continuous $\forall x \in \mathbb{R}$