a) lim _(x arrow 1)(x^2-1) sin (1)/(x-1)=

To evaluate the limit \(\lim_{x \rightarrow 1} (x^2 - 1) \sin \frac{1}{x - 1}\), we can use the substitution \( t = \frac{1}{x - 1} \). As \( x \rightarrow 1 \), \( t \rightarrow \infty \). Therefore, the limit becomes:<br /><br />\[<br />\lim_{t \rightarrow \infty} \left( \left(1 + \frac{1}{t}\right)^2 - 1 \right) \sin(t)<br />\]<br /><br />First, let's simplify the expression inside the parentheses:<br /><br />\[<br />\left(1 + \frac{1}{t}\right)^2 - 1 = 1 + \frac{2}{t} + \frac{1}{t^2} - 1 = \frac{2}{t} + \frac{1}{t^2}<br />\]<br /><br />So the limit becomes:<br /><br />\[<br />\lim_{t \rightarrow \infty} \left( \frac{2}{t} + \frac{1}{t^2} \right) \sin(t)<br />\]<br /><br />As \( t \rightarrow \infty \), both \(\frac{2}{t}\) and \(\frac{1}{t^2}\) approach 0. Therefore, the expression simplifies to:<br /><br />\[<br />\lim_{t \rightarrow \infty} 0 \cdot \sin(t) = 0<br />\]<br /><br />Thus, the limit is:<br /><br />\[<br />\boxed{0}<br />\]