91Ó°ÊÓ

We have been given a function $f\left(x\right)=\left\{\begin{array}{l}\frac{x^3+3x}{x^2−3x}â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x≠0\\ 1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x=0\end{array}\right.$.

We have to determine whether this function is continuous at $c=0$.

We say that f is continuous at c if it is defined at c, given that c is in the domain of f so that f(c) is equal to a number.

Also, another condition is when the right and left limits at c of the function f(x) are both equal to f(c).

$\underset{x→c}{\lim }f\left(x\right)=f\left(c\right)$

However, in a case of a piecewise-defined function, different functions for different intervals must be taken into consideration.

Since $f\left(x\right)=\left\{\begin{array}{l}\frac{x^3+3x}{x^2−3x}â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x≠0\\ 1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x=0\end{array}\right.$

we get$f\left(0\right)=1$

$$\begin{gathered} \underset{x→0^{−}}{lim} \left(\frac{x^3+3x}{x^2−3x}\right) \\ =\underset{x→0^{−}}{lim} \left(\frac{x\left(x^2+3\right)}{x(x−3)}\right) \\ =\underset{x→0^{−}}{lim} \left(\frac{x^2+3}{x−3}\right) \\ =\frac{\underset{x→0^{−}}{lim} x^2+\underset{x→0^{−}}{lim} 3}{\underset{x→0^{−}}{lim} x−\underset{x→0^{−}}{lim} } \\ =\frac{0^2+3}{0−3} \\ =\frac{3}{−3} \\ =-1 \end{gathered}$$

_{$$\begin{gathered} \underset{x→0^{+}}{lim} \left(\frac{x^3+3x}{x^2−3x}\right) \\ =\underset{x→0^{+}}{lim} \left(\frac{x\left(x^2+3\right)}{x(x−3)}\right) \\ =\underset{x→0^{+}}{lim} \left(\frac{x^2+3}{x−3}\right) \\ =\frac{\underset{x→0^{+}}{lim} x^2+\underset{x→0^{+}}{lim} 3}{\underset{x→0^{+}}{lim} x−\underset{x→0^{+}}{lim} } \\ =\frac{0^2+3}{0−3} \\ =\frac{3}{−3} \\ =-1 \end{gathered}$$}

$\underset{x→0}{lim}f\left(x\right)=-1$

Since, $\underset{x→0}{lim}f\left(x\right)≠f\left(0\right)$

The function is not continuous at $c=0$.