It looks like the limit is
[tex]\displaystyle \lim_{x\to3} \frac{x^2 + x - 12}{x^2 - 3x}[/tex]
Factorize the numerator and denominator to reveal a removable discontinuity and evaluate the limit.
[tex]\displaystyle \lim_{x\to3} \frac{x^2 + x - 12}{x^2 - 3x} = \lim_{x\to3} \frac{(x + 4) (x - 3)}{x (x - 3)} = \lim_{x\to3} \frac{x+4}x = \frac{3+4}3 = \boxed{\frac73}[/tex]
