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Let's solve each limit in order.

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### 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,
\[
\lim_{{x \to 0}} \frac{{x - 1}}{{\ln x}} = 0
\]

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### 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) ≈

Evaluate the following limits:

1. \( \lim_{{x \to 0}} \frac{{x - 1}}{{\ln x}} \)
2. \( \lim_{{x \to 1}} \frac{{\sin(\pi x)}}{{x - 1}} \)

Detailed solutions for limit problems involving logarithmic and trigonometric functions. Solve \( \lim_{{x \to 0}} \frac{{x - 1}}{{\ln x}} \) and \( \lim_{{x \to 1}} \frac{{\sin(\pi x)}}{{x - 1}} \) using L'Hôpital's Rule. Suitable for advanced calculus students.

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