91Ó°ÊÓ

We have been given a function $f\left(x\right)=\left\{\begin{array}{l}\frac{x^2−6x}{x^2+6x}â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x≠0\\ −1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x=0\end{array}\right.$.

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$f\left(x\right)=\left\{\begin{array}{l}\frac{x^2−6x}{x^2+6x}â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x≠0\\ −1â€\dots â\P Ä\dots â\P Ä\dots â\P Ä\dots \text{if}x=0\end{array}\right.$

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

$$\begin{gathered} \underset{x→0^{−}}{lim} \left(\frac{x^2−6x}{x^2+6x}\right) \\ =\underset{x→0^{−}}{lim} \left(\frac{x(x−6)}{x(x+6)}\right) \\ =\underset{x→0^{−}}{lim} \left(\frac{x−6}{x+6}\right) \\ =\frac{\underset{x→0^{−}}{lim} x−\underset{x→0^{−}}{lim} 6}{\underset{x→0^{−}}{lim} x+\underset{x→0^{−}}{lim} 6} \\ =\frac{0−6}{0+6} \\ =\frac{−6}{6} \\ =-1 \end{gathered}$$

$$\begin{gathered} \underset{x→0^{+}}{lim} \left(\frac{x^2−6x}{x^2+6x}\right) \\ =\underset{x→0^{+}}{lim} \left(\frac{x(x−6)}{x(x+6)}\right) \\ =\underset{x→0+}{lim} \left(\frac{x−6}{x+6}\right) \\ =\frac{\underset{x→0^{+}}{lim} x−\underset{x→0^{+}}{lim} 6}{\underset{x→0^{+}}{lim} x+\underset{x→0^{+}}{lim} 6} \\ =\frac{0−6}{0+6} \\ =\frac{−6}{6} \\ =-1 \end{gathered}$$

localid="1647085301117" $$\begin{gathered} \underset{x→0^{-}}{\lim }f\left(x\right)=\underset{x→0^{+}}{\lim }f\left(x\right) \\ \underset{x→0}{\lim }f\left(x\right)=-1 \end{gathered}$$

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

The function is continuous at $c=0$.