To find the limit [tex]\(\lim_{x \rightarrow 3}\left(\frac{1}{x-3} - \frac{6}{x^2 - 9}\right)\)[/tex], let's go through a detailed step-by-step solution.
1. Understanding the components:
- Notice that [tex]\(x^2 - 9\)[/tex] is a difference of squares and can be factored as [tex]\( (x-3)(x+3) \)[/tex].
- Therefore, [tex]\( \frac{6}{x^2 - 9} \)[/tex] can be written as [tex]\( \frac{6}{(x-3)(x+3)} \)[/tex].
2. Rewrite the limit expression correctly:
[tex]\[
\lim_{x \rightarrow 3}\left(\frac{1}{x-3} - \frac{6}{(x-3)(x+3)}\right)
\][/tex]
3. Combine the fractions:
- To combine these fractions into a single fraction, find a common denominator. The common denominator between [tex]\( x-3 \)[/tex] and [tex]\( (x-3)(x+3) \)[/tex] is [tex]\( (x-3)(x+3) \)[/tex].
- Write each fraction with the common denominator:
[tex]\[
\frac{1}{x-3} = \frac{(x+3)}{(x-3)(x+3)}
\][/tex]
and
[tex]\[
\frac{6}{(x-3)(x+3)} \text{ is already with the common denominator}.
\][/tex]
- Now, rewrite the original expression with the common denominator:
[tex]\[
\lim_{x \rightarrow 3} \left(\frac{x+3}{(x-3)(x+3)} - \frac{6}{(x-3)(x+3)}\right)
\][/tex]
Simplify the numerator:
[tex]\[
\lim_{x \rightarrow 3} \left(\frac{x+3 - 6}{(x-3)(x+3)}\right)
\][/tex]
[tex]\[
\lim_{x \rightarrow 3} \left(\frac{x-3}{(x-3)(x+3)}\right)
\][/tex]
Cancelling the common factor [tex]\( x-3 \)[/tex] from the numerator and the denominator:
[tex]\[
\lim_{x \rightarrow 3} \left(\frac{1}{x+3}\right)
\][/tex]
4. Evaluate the limit:
- Now, as [tex]\( x \)[/tex] approaches 3, substitute [tex]\( x \)[/tex] with 3 in the simplified expression:
[tex]\[
\frac{1}{3+3} = \frac{1}{6}
\][/tex]
5. Result:
[tex]\[
\boxed{\frac{1}{6}}
\][/tex]
Thus, the limit [tex]\(\lim _{x \rightarrow 3}\left(\frac{1}{x-3} - \frac{6}{x^2-9}\right)\)[/tex] is [tex]\(\frac{1}{6}\)[/tex].
