Question

part 1: evaluate the limit. evaluate the following limit by simplifying the expression (first answer - box) and then evaluating the limit (second answer - box). (lim_{x
ightarrow8}\frac{x - 8}{sqrt{x}-sqrt{8}}=lim_{x
ightarrow8}square=square). hint: treat (x - 8) as a difference of squares. note: in your written solution, you should write the limit statement (lim_{x
ightarrow8}) in every step except the last one, where the limit is finally evaluated. part 2: follow - up question

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt{x}+\sqrt{8}$.
\[

$$\begin{align*} \lim_{x ightarrow8}\frac{x - 8}{\sqrt{x}-\sqrt{8}}&=\lim_{x ightarrow8}\frac{(x - 8)(\sqrt{x}+\sqrt{8})}{(\sqrt{x}-\sqrt{8})(\sqrt{x}+\sqrt{8})}\\ \end{align*}$$

\]
Since $(a - b)(a + b)=a^{2}-b^{2}$, the denominator $(\sqrt{x}-\sqrt{8})(\sqrt{x}+\sqrt{8})=x - 8$.
So the expression becomes $\lim_{x
ightarrow8}\frac{(x - 8)(\sqrt{x}+\sqrt{8})}{x - 8}$.

Step2: Simplify the expression

Cancel out the common factor $(x - 8)$ (for $x
eq8$).
\[
\lim_{x
ightarrow8}\frac{(x - 8)(\sqrt{x}+\sqrt{8})}{x - 8}=\lim_{x
ightarrow8}(\sqrt{x}+\sqrt{8})
\]

Step3: Evaluate the limit

Substitute $x = 8$ into $\sqrt{x}+\sqrt{8}$.
\[
\lim_{x
ightarrow8}(\sqrt{x}+\sqrt{8})=\sqrt{8}+\sqrt{8}=2\sqrt{8}=4\sqrt{2}
\]

Answer:

$4\sqrt{2}$