To estimate the limit [tex]\(\lim_{{x \to 3}} \frac{x-3}{x^2 - 2x - 3}\)[/tex], let's go through a detailed step-by-step solution.
1. Identify the expression in the limit:
[tex]\[
\lim_{{x \to 3}} \frac{x-3}{x^2 - 2x - 3}
\][/tex]
2. Factor the denominator:
The quadratic expression [tex]\(x^2 - 2x - 3\)[/tex] can be factored. We find two numbers that multiply to [tex]\(-3\)[/tex] and add to [tex]\(-2\)[/tex]: these numbers are [tex]\(1\)[/tex] and [tex]\(-3\)[/tex].
[tex]\[
x^2 - 2x - 3 = (x - 3)(x + 1)
\][/tex]
3. Rewrite the original limit with the factored form:
[tex]\[
\lim_{{x \to 3}} \frac{x-3}{(x - 3)(x + 1)}
\][/tex]
4. Simplify the expression:
Cancel out the common factor [tex]\((x - 3)\)[/tex] from the numerator and denominator, provided [tex]\(x \neq 3\)[/tex]:
[tex]\[
\frac{x-3}{(x - 3)(x + 1)} = \frac{1}{x + 1}
\][/tex]
5. Consider the simplified expression:
Now, the limit becomes:
[tex]\[
\lim_{{x \to 3}} \frac{1}{x + 1}
\][/tex]
6. Substitute [tex]\(x = 3\)[/tex] in the simplified expression:
[tex]\[
\frac{1}{3 + 1} = \frac{1}{4}
\][/tex]
Therefore, the limit is:
[tex]\[
\lim_{{x \to 3}} \frac{x-3}{x^2 - 2x - 3} = \frac{1}{4} = 0.25
\][/tex]
Given the options:
a. [tex]\(\lim_{{x \to 3}} f(x) = 0.2\)[/tex]
b. [tex]\(\lim_{{x \to 3}} f(x) = 0.24\)[/tex]
c. [tex]\(\lim_{{x \to 3}} f(x) = 0.25\)[/tex]
d. [tex]\(\lim_{{x \to 3}} f(x) = 0.21\)[/tex]
The correct answer is:
c. [tex]\(\lim_{{x \to 3}} f(x) = 0.25\)[/tex]
