Answer
Let $f(x)=\displaystyle \frac{x^{2}-7x+12}{x-3}$ and $g(x)=(x-4)$
Since $\displaystyle \frac{x^{2}-7x+12}{x-3}=\frac{(x-3)(x-4)}{(x-3)}$ it follows that,
for all x except 3, the factors $(x-3)$ cancel, and $f(x)=g(x)$ whenever $x\neq 3$.
Since g is a linear function, by th.2.2 it has a limit at x=3.
By the previous exercise, it follows that $\displaystyle \lim_{x\rightarrow 3}f(x)=\lim_{x\rightarrow 3}g(x).$
Work Step by Step
The explanation is given in the answer to the question of the exercise.
