Let's find the limit [tex]\(\lim _{x \rightarrow a} \frac{\sin (x-a)}{x^2-a^2}\)[/tex].
### Step-by-Step Solution:
1. Understand the form
Notice that as [tex]\( x \)[/tex] approaches [tex]\( a \)[/tex], both the numerator and the denominator approach 0, making this an indeterminate form of [tex]\(\frac{0}{0}\)[/tex]. This suggests that L'Hôpital's Rule might be suitable for solving it.
2. Apply L'Hôpital's Rule
L'Hôpital's Rule states that if [tex]\( \lim_{x \to c} \frac{f(x)}{g(x)} \)[/tex] results in an indeterminate form [tex]\( \frac{0}{0} \)[/tex] or [tex]\( \frac{\infty}{\infty} \)[/tex], then:
[tex]\[
\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)},
\][/tex]
given that the limits of the derivatives exist.
3. Differentiate the numerator and denominator
[tex]\[
f(x) = \sin(x - a) \quad \Rightarrow \quad f'(x) = \cos(x - a)
\][/tex]
[tex]\[
g(x) = x^2 - a^2 = (x - a)(x + a) \quad \Rightarrow \quad g'(x) = 2x
\][/tex]
4. Rewrite the limit using the derivatives
Thus, applying L'Hôpital's Rule gives us:
[tex]\[
\lim _{x \rightarrow a} \frac{\sin(x - a)}{x^2 - a^2} = \lim _{x \rightarrow a} \frac{\cos(x - a)}{2x}
\][/tex]
5. Evaluate the limit
Substitute [tex]\( x = a \)[/tex] into the new expression:
[tex]\[
\lim _{x \rightarrow a} \frac{\cos(x - a)}{2x} = \frac{\cos(a - a)}{2a} = \frac{\cos(0)}{2a}
\][/tex]
Since [tex]\(\cos(0) = 1\)[/tex]:
[tex]\[
\frac{1}{2a}
\][/tex]
So, the limit is:
[tex]\[
\lim _{x \rightarrow a} \frac{\sin (x-a)}{x^2-a^2} = \frac{1}{2a}
\][/tex]
This completes our solution.
