(a) \( \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3} \)

To evaluate the limit \( \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3} \), we first consider what happens as \( x \) approaches 3 from the right (values slightly greater than 3). The expression can be rewritten as: \[ \frac{x-1}{x-3} \] As \( x \) approaches 3, the numerator \( x - 1 \) approaches 2 (since \( 3 - 1 = 2 \)), and the denominator \( x - 3 \) approaches 0 from the positive side (since we are approaching from the right, \( x - 3 \) will be a small positive number). Thus, we have: \[ \lim _{x \rightarrow 3^{+}} \frac{x-1}{x-3} = \frac{2}{0^{+}} = +\infty \] So, the limit is \( +\infty \).