Question

evaluate the limit:
$lim_{x
ightarrow6}\frac{x - 6}{x^{2}-6x}=square$
enter dne if the limit does not exist.

Explanation:

Step1: Factor the denominator

Factor $x^{2}-6x$ as $x(x - 6)$. So the limit becomes $\lim_{x
ightarrow6}\frac{x - 6}{x(x - 6)}$.

Step2: Simplify the function

Cancel out the common factor $(x - 6)$ (since $x
eq6$ when taking the limit), we get $\lim_{x
ightarrow6}\frac{1}{x}$.

Step3: Substitute the value of $x$

Substitute $x = 6$ into $\frac{1}{x}$, we have $\frac{1}{6}$.

Answer:

$\frac{1}{6}$