Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{x
ightarrow36}\frac{36 - x}{6-sqrt{x}}

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $6 + \sqrt{x}$.
\[

$$\begin{align*} \lim_{x ightarrow36}\frac{36 - x}{6-\sqrt{x}}&=\lim_{x ightarrow36}\frac{(36 - x)(6+\sqrt{x})}{(6-\sqrt{x})(6 + \sqrt{x})}\\ \end{align*}$$

\]
Since $(a - b)(a + b)=a^{2}-b^{2}$, the denominator is $36 - x$.
So the expression becomes $\lim_{x
ightarrow36}(6+\sqrt{x})$.

Step2: Substitute the value of $x$

Substitute $x = 36$ into $6+\sqrt{x}$.
When $x = 36$, $\sqrt{x}=6$, and $6+\sqrt{x}=6 + 6=12$.

Answer:

$12$