To solve the limit [tex]\(\lim_{x \rightarrow 3} \left(\frac{1}{x-3} - \frac{6}{x^2 - 9}\right)\)[/tex], follow these steps:
1. Simplify the Expression:
First, simplify the second term [tex]\(\frac{6}{x^2 - 9}\)[/tex]. Notice that [tex]\(x^2 - 9\)[/tex] can be factored using the difference of squares:
[tex]\[
x^2 - 9 = (x - 3)(x + 3)
\][/tex]
So, the expression becomes:
[tex]\[
\frac{1}{x-3} - \frac{6}{(x-3)(x+3)}
\][/tex]
2. Combine the Fractions:
To combine these fractions, find a common denominator. The common denominator for [tex]\(\frac{1}{x-3}\)[/tex] and [tex]\(\frac{6}{(x-3)(x+3)}\)[/tex] is [tex]\((x-3)(x+3)\)[/tex]:
[tex]\[
\left( \frac{1}{x-3} - \frac{6}{(x-3)(x+3)} \right)
= \frac{(x+3) - 6}{(x-3)(x+3)}
\][/tex]
3. Simplify the Numerator:
Simplify the numerator:
[tex]\[
(x+3) - 6 = x - 3
\][/tex]
So, the expression now is:
[tex]\[
\frac{x - 3}{(x - 3)(x + 3)}
\][/tex]
4. Cancel the Common Factor:
Cancel the [tex]\((x - 3)\)[/tex] term in the numerator and the denominator:
[tex]\[
\frac{x-3}{(x-3)(x+3)} = \frac{1}{x+3} \quad \text{for} \quad x \neq 3
\][/tex]
5. Evaluate the Limit:
Now, we need to find the limit of [tex]\(\frac{1}{x+3}\)[/tex] as [tex]\( x \)[/tex] approaches 3:
[tex]\[
\lim_{x \to 3} \frac{1}{x + 3}
\][/tex]
6. Substitute [tex]\(x = 3\)[/tex]:
Simply substitute [tex]\(x = 3\)[/tex] in [tex]\(\frac{1}{x+3}\)[/tex]:
[tex]\[
\frac{1}{3+3} = \frac{1}{6}
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \rightarrow 3} \left( \frac{1}{x-3} - \frac{6}{x^2-9} \right) = \frac{1}{6}
\][/tex]
