evaluate the limit\n$lim_{b\rightarrow81}\frac{81 - b}{9-sqrt{b}}=square$.
evaluate the limit\n$lim_{b\rightarrow81}\frac{81 - b}{9-sqrt{b}}=square$.
Answer
Explanation:
Step1: Rationalize the denominator
Multiply the numerator and denominator by the conjugate of the denominator $9 + \sqrt{b}$. [ \begin{align*} \lim_{b\rightarrow81}\frac{81 - b}{9-\sqrt{b}}&=\lim_{b\rightarrow81}\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\rightarrow81}\frac{(81 - b)(9+\sqrt{b})}{81 - b}\ \end{align*} ]
Step2: Simplify the expression
Cancel out the common factor $81 - b$ (since $b\neq81$ when taking the limit). [ \begin{align*} &=\lim_{b\rightarrow81}(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:
$18$