Question

How to evaluate the limit

Answer 1

$\displaystyle\lim_{x\to2}\frac{x^2-x+6}{x+2}$

Both the numerator and denominator are continuous at $x=2$, which means the quotient rule for limits applies:

$\dfrac{\displaystyle\lim_{x\to2}(x^2-x+6)}{\displaystyle\lim_{x\to2}(x+2)}=\dfrac{2^2-2+6}{2+2}=\dfrac84=2$

Perhaps you meant to write that $x\to-2$ instead? In that case, you would have

$\displaystyle\lim_{x\to-2}\frac{x^2-x+6}{x+2}=\lim_{x\to-2}\frac{(x+2)(x-3)}{x+2}=\lim_{x\to-2}(x-3)=-2-3=-5$