2) \( \lim _{x \rightarrow 1} \frac{\sin (x-1)}{5\left(x^{2}-1\right)} \)

To evaluate the limit \( \lim _{x \rightarrow 1} \frac{\sin (x-1)}{5\left(x^{2}-1\right)} \), we start by observing that direct substitution of \( x = 1 \) results in the indeterminate form \( \frac{0}{0} \). To resolve this, we can apply L'Hôpital's Rule, which states that we can take the derivatives of the numerator and denominator. The derivative of the numerator \( \sin(x-1) \) is \( \cos(x-1) \), and the derivative of the denominator \( 5(x^2 - 1) \) is \( 10x \). Thus, we get: \[ \lim _{x \rightarrow 1} \frac{\cos (x-1)}{10x} \] Now substitute \( x = 1 \) into the new limit: \[ \frac{\cos(1-1)}{10 \cdot 1} = \frac{\cos(0)}{10} = \frac{1}{10} \] Therefore, the limit is: \[ \boxed{\frac{1}{10}} \]