Question

evaluate the limit
$lim_{b
ightarrow81}\frac{81 - b}{9-sqrt{b}}=square$.

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by the conjugate of the denominator $9 + \sqrt{b}$.
\[

$$\begin{align*} \lim_{b ightarrow81}\frac{81 - b}{9-\sqrt{b}}&=\lim_{b ightarrow81}\frac{(81 - b)(9+\sqrt{b})}{(9-\sqrt{b})(9 + \sqrt{b})}\\ \end{align*}$$

\]
Since $(a - b)(a + b)=a^{2}-b^{2}$, the denominator is $81 - b$.
\[

$$\begin{align*} &=\lim_{b ightarrow81}\frac{(81 - b)(9+\sqrt{b})}{81 - b}\\ \end{align*}$$

\]

Step2: Simplify the expression

Cancel out the common factor $81 - b$ (since $b
eq81$ when taking the limit).
\[

$$\begin{align*} &=\lim_{b ightarrow81}(9+\sqrt{b}) \end{align*}$$

\]

Step3: Evaluate the limit

Substitute $b = 81$ into the expression.
\[

$$\begin{align*} &=9+\sqrt{81}\\ &=9 + 9\\ &=18 \end{align*}$$

\]

Answer:

$18$