The function is continuous if f(c) is defined, where c is a constant and the limit of x approach to c give the same value of f(c)
For the given function
[tex]f(x)=\frac{x^2}{x-3}[/tex]
If f(-3) is defined, then the function is continuous at this value
Let us find f(-3) by substituting x by -3
[tex]\begin{gathered} f(-3)=\frac{(-3)^2}{-3-3} \\ f(-3)=\frac{9}{-6} \\ f(-3)=-\frac{3}{2} \end{gathered}[/tex]
Then f(-3) = -3/2
[tex]\lim _{x\rightarrow-3}\frac{x^2}{x-3}=\frac{(-3)^2}{-3-3}=\frac{9}{-6}=-\frac{3}{2}[/tex]
Since f(-3) has the same value of the limit of x approach to -3, then
The function is continuous at x = -3
