The function is
[tex]f(x)=\frac{(x-2)^2(x+1)(x+3)}{x^3(x+3)(x-4)}[/tex]
Notice that if we try to evaluate the function at x=0, we would get,
[tex]\lim _{x\to0}f(x)=\frac{12}{0}[/tex]
This means that the function f(x) goes to infinity as x approaches zero.
Consider a very small value ε>0 and calculate the f(x) limit when x approaches ε.
[tex]\lim _{x\to\epsilon}f(x)=\frac{(\epsilon-2)^2(\epsilon+1)(\epsilon+3)}{\epsilon^3(\epsilon+3)(\epsilon-4)}[/tex]
Notice that the numerator is positive for any small value of ε>0. On the other hand, regarding the denominator (ε^3)>0, (ε+3)>0, (ε-4)<0. There is a negative value in the denominator!
Therefore, the function approaches -infinite as x->0 from the right
[tex]\lim _{x\to0^+}f(x)=-\infty[/tex]
