To evaluate the limit [tex]\(\lim_{x \rightarrow 9} \frac{x-9}{\sqrt{x}-3}\)[/tex], we will proceed with the following steps:
1. Direct Substitution: First, try substituting [tex]\( x = 9 \)[/tex] directly into the expression to see if it results in a determinate form.
[tex]\[
\frac{9-9}{\sqrt{9}-3} = \frac{0}{0}
\][/tex]
This gives us the indeterminate form [tex]\(\frac{0}{0}\)[/tex].
2. Rewrite the Expression: To resolve the indeterminate form, we need to manipulate the expression algebraically. Notice that we can factor and simplify the expression. Start by rewriting the numerator:
[tex]\[
x - 9 = (\sqrt{x})^2 - 3^2
\][/tex]
Recognize that this is a difference of squares, so we can factor it as follows:
[tex]\[
x - 9 = (\sqrt{x} - 3)(\sqrt{x} + 3)
\][/tex]
3. Simplify the Fraction: Substitute this factorization back into the original limit:
[tex]\[
\frac{x - 9}{\sqrt{x} - 3} = \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3}
\][/tex]
Since [tex]\(\sqrt{x} \neq 3\)[/tex] for [tex]\(x\)[/tex] close to (but not equal to) 9, we can cancel out the common factor [tex]\(\sqrt{x} - 3\)[/tex] in the numerator and the denominator:
[tex]\[
\frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{\sqrt{x} - 3} = \sqrt{x} + 3
\][/tex]
4. Evaluate the Limit: Now that we have simplified the expression, we can directly substitute [tex]\( x = 9 \)[/tex]:
[tex]\[
\lim_{x \to 9} (\sqrt{x} + 3) = \sqrt{9} + 3
\][/tex]
[tex]\[
= 3 + 3
\][/tex]
[tex]\[
= 6
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{x \rightarrow 9} \frac{x-9}{\sqrt{x}-3} = 6
\][/tex]
