Let's solve each limit in order.

Problem 120:

$\lim_{{x \to 0}} \frac{{x - 1}}{{\ln x}}$

Solution:

As $x \to 0$, $\ln x \to -\infty$ and $x - 1 \to -1$, resulting in an indeterminate form of $\frac{-1}{-\infty} = 0$.

Thus, $\lim_{{x \to 0}} \frac{{x - 1}}{{\ln x}} = 0$

Problem 121:

$\lim_{{x \to 1}} \frac{{\sin (\pi x)}}{{x - 1}}$

Solution:

We can rewrite $\sin (\pi x)$ as $\sin (\pi (x - 1 + 1)) = \sin (\pi (x - 1) + \pi)$.

Using the small-angle approximation, (\sin (\pi (x - 1) + \pi) ≈