To find the limit
[tex]\[
\lim _{x \rightarrow 81} \frac{\sqrt{x}-9}{x-81},
\][/tex]
we need to analyze the behavior of the function [tex]\(\frac{\sqrt{x}-9}{x-81}\)[/tex] as [tex]\(x\)[/tex] approaches 81.
First, let's recognize that direct substitution of [tex]\(x = 81\)[/tex] into the function results in an indeterminate form:
[tex]\[
\frac{\sqrt{81} - 9}{81 - 81} = \frac{9 - 9}{0} = \frac{0}{0}.
\][/tex]
Given the indeterminate form [tex]\(\frac{0}{0}\)[/tex], we can apply algebraic techniques to simplify the expression.
Observe that:
[tex]\[
\sqrt{x} - 9 = (\sqrt{x} - 9)
\][/tex]
and [tex]\(x - 81\)[/tex] can be factored further. Since [tex]\(81 = 9^2\)[/tex], we use conjugates to simplify the fraction. Multiply the numerator and the denominator by the conjugate of the numerator, [tex]\(\sqrt{x} + 9\)[/tex]:
[tex]\[
\frac{\sqrt{x} - 9}{x - 81} \cdot \frac{\sqrt{x} + 9}{\sqrt{x} + 9} = \frac{(\sqrt{x} - 9)(\sqrt{x} + 9)}{(x - 81)(\sqrt{x} + 9)}.
\][/tex]
Now, use the difference of squares to simplify the numerator:
[tex]\[
(\sqrt{x} - 9)(\sqrt{x} + 9) = x - 81.
\][/tex]
So the fraction becomes:
[tex]\[
\frac{x - 81}{(x - 81)(\sqrt{x} + 9)} = \frac{1}{\sqrt{x} + 9}.
\][/tex]
Now, take the limit as [tex]\(x \rightarrow 81\)[/tex]:
[tex]\[
\lim_{x \rightarrow 81} \frac{1}{\sqrt{x} + 9}.
\][/tex]
Substitute [tex]\(x = 81\)[/tex] into the simplified expression:
[tex]\[
\frac{1}{\sqrt{81} + 9} = \frac{1}{9 + 9} = \frac{1}{18}.
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \rightarrow 81} \frac{\sqrt{x} - 9}{x - 81} = \frac{1}{18} = 0.05555555555555555.
\][/tex]
Rounded to at least three decimal places, the answer is [tex]\(0.0555\)[/tex].
