Question

1. - / 3.84 points evaluate the limit, if it exists. (if an answer does not exist, enter dne.) $lim_{x

ightarrow6}\frac{x^{2}-5x + 6}{x - 6}$

Explanation:

Step1: Factor the numerator

Factor $x^{2}-5x + 6$. We know that $x^{2}-5x + 6=(x - 2)(x - 3)$. So the limit becomes $\lim_{x
ightarrow6}\frac{(x - 2)(x - 3)}{x - 6}$.

Step2: Substitute $x = 6$

If we directly substitute $x = 6$ into $\frac{(x - 2)(x - 3)}{x - 6}$, we get $\frac{(6 - 2)(6 - 3)}{6 - 6}=\frac{4\times3}{0}$, which is undefined.

Step3: Simplify the function

Since the original function has a zero - denominator when $x = 6$, we can't directly evaluate the limit in this form. But we can rewrite the limit as follows:
\[

$$\begin{align*} \lim_{x ightarrow6}\frac{x^{2}-5x + 6}{x - 6}&=\lim_{x ightarrow6}\frac{(x - 2)(x - 3)}{x - 6}\\ \end{align*}$$

\]
As $x
ightarrow6$, the numerator approaches $(6 - 2)(6 - 3)=4\times3 = 12$ and the denominator approaches $0$. The left - hand limit $\lim_{x
ightarrow6^{-}}\frac{x^{2}-5x + 6}{x - 6}=-\infty$ and the right - hand limit $\lim_{x
ightarrow6^{+}}\frac{x^{2}-5x + 6}{x - 6}=\infty$.

Answer:

DNE