Answer
$-3$
Work Step by Step
We want to find $\lim\limits_{x \to 2}\dfrac{x^2-7x+10}{x-2}$, but we can't use the quotient rule for limits because the limit of the denominator as $x$ approaches 2 is zero, and if we try to substitute $x=2$ directly, we get zero in both the numerator and the denominator. This means we have a common factor of $x-2$ in both the numerator and the denominator.
Note that $\dfrac{x^2-7x+10}{x-2}=\dfrac{(x-2)(x-5)}{x-2}=x-5$ for all $x \neq 2.$
Thus $\lim\limits_{x \to 2}\dfrac{x^2-7x+10}{x-2}=\lim\limits_{x \to 2}x-5=2-5=-3.$
