To find the limit [tex]\(\lim_{x \rightarrow a} \frac{\sin (x-a)}{x^2 - a^2}\)[/tex], let's proceed step by step.
First, observe the form of the expression when [tex]\(x\)[/tex] approaches [tex]\(a\)[/tex]:
[tex]\[
\lim_{x \rightarrow a} \frac{\sin (x-a)}{x^2 - a^2}
\][/tex]
If we directly substitute [tex]\(x = a\)[/tex] into the expression, we get:
[tex]\[
\frac{\sin(a-a)}{a^2 - a^2} = \frac{0}{0}
\][/tex]
This is an indeterminate form, so we must simplify or transform the expression to evaluate the limit.
Notice that [tex]\(x^2 - a^2\)[/tex] can be factored using the difference of squares:
[tex]\[
x^2 - a^2 = (x-a)(x+a)
\][/tex]
So we can rewrite the limit as:
[tex]\[
\lim_{x \rightarrow a} \frac{\sin (x-a)}{(x-a)(x+a)}
\][/tex]
Now, we can separate the fraction:
[tex]\[
\lim_{x \rightarrow a} \frac{\sin (x-a)}{x-a} \cdot \frac{1}{x+a}
\][/tex]
Let's address each part separately. Consider the limit of the first factor. We know a standard limit result:
[tex]\[
\lim_{t \rightarrow 0} \frac{\sin t}{t} = 1
\][/tex]
By making the substitution [tex]\(t = x - a\)[/tex], as [tex]\(x\)[/tex] approaches [tex]\(a\)[/tex], [tex]\(t\)[/tex] approaches [tex]\(0\)[/tex]. Therefore:
[tex]\[
\lim_{x \rightarrow a} \frac{\sin (x-a)}{x-a} = 1
\][/tex]
Next, consider the second factor. As [tex]\(x \rightarrow a\)[/tex], the term [tex]\((x+a)\)[/tex] becomes [tex]\((a+a) = 2a\)[/tex]:
[tex]\[
\lim_{x \rightarrow a} \frac{1}{x+a} = \frac{1}{2a}
\][/tex]
Combining these two results, we have:
[tex]\[
\lim_{x \rightarrow a} \frac{\sin (x-a)}{(x-a)(x+a)} = \left( \lim_{x \rightarrow a} \frac{\sin (x-a)}{x-a} \right) \cdot \left( \lim_{x \rightarrow a} \frac{1}{x+a} \right)
\][/tex]
[tex]\[
= 1 \cdot \frac{1}{2a} = \frac{1}{2a}
\][/tex]
Thus, the limit is:
[tex]\[
\boxed{\frac{1}{2a}}
\][/tex]
