To solve the limit [tex]\(\lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^2-x-2}\)[/tex], we will follow a systematic approach. Let’s break it down step-by-step:
1. Identify the nature of the function:
We need to evaluate [tex]\(\frac{\sin (x-2)}{x^2 - x - 2}\)[/tex] as [tex]\(x\)[/tex] approaches 2. First, let's understand each component separately:
- The numerator: [tex]\(\sin(x-2)\)[/tex]
- The denominator: [tex]\(x^2 - x - 2\)[/tex]
2. Factorize the denominator:
Factorize [tex]\(x^2 - x - 2\)[/tex]. This trinomial can be factored into:
[tex]\[
x^2 - x - 2 = (x-2)(x+1)
\][/tex]
So, the original function can be rewritten as:
[tex]\[
\frac{\sin(x-2)}{(x-2)(x+1)}
\][/tex]
3. Simplify the expression:
Notice that the term [tex]\(x-2\)[/tex] in the numerator can potentially cancel with the term [tex]\(x-2\)[/tex] in the denominator. However, this cancellation is valid only if [tex]\(x \neq 2\)[/tex]. So we rewrite the function as:
[tex]\[
\frac{\sin(x-2)}{(x-2)(x+1)} = \frac{1}{x+1} \cdot \frac{\sin(x-2)}{x-2}
\][/tex]
4. Evaluate the limit:
Evaluate each part of the product separately:
- [tex]\(\lim_{x \to 2} \frac{\sin(x-2)}{x-2}\)[/tex]
- [tex]\(\lim_{x \to 2} \frac{1}{x+1}\)[/tex]
Recall the standard limit result:
[tex]\[
\lim_{y \to 0} \frac{\sin y}{y} = 1
\][/tex]
Here, we substitute [tex]\(y = x - 2\)[/tex], which implies [tex]\(y\)[/tex] approaches 0 as [tex]\(x\)[/tex] approaches 2. Thus:
[tex]\[
\lim_{x \to 2} \frac{\sin(x-2)}{x-2} = 1
\][/tex]
Next, evaluate the simpler part:
[tex]\[
\lim_{x \to 2} \frac{1}{x+1} = \frac{1}{2+1} = \frac{1}{3}
\][/tex]
5. Combine the results:
Multiplying both limits together gives:
[tex]\[
\lim_{x \to 2} \frac{\sin(x-2)}{(x-2)(x+1)} = \left( \lim_{x \to 2} \frac{\sin(x-2)}{x-2} \right) \cdot \left( \lim_{x \to 2} \frac{1}{x+1} \right)
\][/tex]
[tex]\[
= 1 \cdot \frac{1}{3} = \frac{1}{3}
\][/tex]
Therefore, the limit is:
[tex]\[
\boxed{\frac{1}{3}}
\][/tex]
