Example 3. Find $\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x-4}$.
Solution. When $x=4$, both the numerator and the denominator of the given function become zero. This results in an indeterminate form $\frac{0}{0}$, which needs to be resolved. We will transform the given function by factoring the numerator using the formula

$$
x^{2}+p x+q=\left(x-x_{1}\right)\left(x-x_{2}\right)
$$

where $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2}+p x+q=0$.

Since the equation $x^{2}-6 x+8=0$ has roots $x_{1}=2, x_{2}=4$, we have $x^{2}-6 x+8=(x-2)(x-4)$. Substituting this expression into the given function and canceling the common factor $(x-4) \neq 0$, we get

$$
\lim _{x \rightarrow 4} \frac{x^{2}-6 x+8}{x-4}=\lim _{x \rightarrow 4} \frac{(x-2)(x-4)}{x-4}=\lim _{x \rightarrow 4}(x-2)=4-2=2
$$

Remark 1. When canceling the common factor, we assumed that $x-4 \neq 0$. This is indeed the case. According to the definition of the limit of a function, the argument $x$ approaches its limiting value $a$ without ever coinciding with it, i.e., $x \neq a$ and $x-a \neq 0$. (See formula (1.26).)

Remark 2. The limit of a function does not depend on whether the function is defined at the limiting point or not.