To estimate the limit [tex]\(\lim _{x \rightarrow 3} \frac{x-3}{x^2-9}\)[/tex], we need to simplify the expression and then evaluate the limit as [tex]\(x\)[/tex] approaches 3.
First, notice that the expression [tex]\(\frac{x-3}{x^2-9}\)[/tex] can be simplified because the denominator [tex]\(x^2 - 9\)[/tex] can be factored:
[tex]\[x^2 - 9 = (x-3)(x+3)\][/tex]
Substituting this into the original expression gives us:
[tex]\[
\frac{x-3}{x^2-9} = \frac{x-3}{(x-3)(x+3)}
\][/tex]
By canceling out the common [tex]\((x-3)\)[/tex] term in the numerator and denominator, the expression simplifies to:
[tex]\[
\frac{1}{x+3}
\][/tex]
Next, we need to find the limit of [tex]\(\frac{1}{x+3}\)[/tex] as [tex]\(x\)[/tex] approaches 3. We substitute [tex]\(x = 3\)[/tex] into the simplified expression:
[tex]\[
\lim_{x \rightarrow 3} \frac{1}{x+3} = \frac{1}{3+3} = \frac{1}{6}
\][/tex]
Therefore, the value of the limit is [tex]\(\frac{1}{6}\)[/tex], which is approximately 0.1667.
From the given choices, the answer that matches this result is:
b. [tex]\(\lim _{x \rightarrow 3} f(x)=0.1667\)[/tex]
So, the best answer is:
b. [tex]\(\lim _{x \rightarrow 3} f(x)=0.1667\)[/tex]
