Answer:
There is a non-removable discontinuity at x = 3
Step-by-step explanation:
We are given that
[tex]f(x)=\frac{x^2+9}{x-3}[/tex]
We have to find true statement about given function.
[tex]\lim_{x\rightarrow 3}f(x)=\lim_{x\rightarrow 3}\frac{x^2+9}{x-3}[/tex]
=[tex]\infty[/tex]
It is not removable discontinuity.
x-3=0
x=3
The function f(x) is not define at x=3. Therefore, the function f(x) is continuous for all real numbers except x=3.
Therefore, x=3 is non- removable discontinuity of function f(x).
Hence, option B is correct.
