https://file.aiquickdraw.com/mathbot/file/1730374991960nvg58sh7.jpg

The problem asks to find the points of discontinuity of the function
\[
f(x) = \begin{cases} 
\frac{x}{\lg(x + 1)}, & x > -1, \\ 
\sin(\pi x), & x \leq -1 
\end{cases}
\]
and determine the type of each discontinuity. Additionally, we are asked to provide a graphical illustration.

### Step 1: Analyze Each Piece of the Function
1. **For \( x > -1 \):** \( f(x) = \frac{x}{\lg(x + 1)} \).
   - This expression is defined when \( x + 1 > 0 \) (so \( x > -1 \)) and \( \lg(x + 1) \neq 0 \).
   - \( \lg(x + 1) = 0 \) when \( x + 1 = 1 \), or \( x = 0 \). At \( x = 0 \), the function may have a discontinuity.

2. **For \( x \leq -1 \):** \( f(x) = \sin(\pi x) \).
   - This function is continuous for all \( x \), including at \( x = -1 \).

### Step 2: Check Discontinuity at \( x = -1 \)
To analyze the continuity at \( x = -1 \), examine the left-hand limit and right-hand limit.

1. **Left-hand limit (as \( x \to -1^- \)):**
   \[
   \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} \sin(\pi x) = \sin(-\pi) = 0.
   \]

2. **Right-hand limit (as \( x \to -1^+ \)):**
   \[
   \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} \frac{x}{\lg(x + 1)}.
   \]
   Here, as \( x \to -1^+ \), \( \lg(x + 1) \to -\infty \) because \( \lg(0^+) = -\infty \), making the fraction \( \frac{x}{\lg(x + 1)} \) approach 0 as well. Thus,
   \[
   \lim_{x \to -1^+} f(x) = 0.
   \]

Since both limits as \( x \to -1 \) yield 0, the function \( f(x) \) is continuous at \( x = -1 \).

### Step 3: Check Discontinuity at \( x = 0 \)
For \( x > -1 \), the function \( f(x) = \frac{x}{\lg(x + 1)} \) has a potential issue at \( x = 0 \) since \( \lg(1) = 0 \), making the denominator zero and the function undefined at \( x = 0 \).

Thus, \( x = 0 \) is a point of discontinuity, specifically a **non-removable discontinuity** (or an infinite discontinuity) because \( f(x) \) becomes unbounded as \( x \) approaches 0.

### Conclusion
- **Discontinuity at \( x = 0 \)**: Non-removable (infinite) discontinuity.
- **No discontinuity at \( x = -1 \)**, as the function is continuous there.

Would you like to see the graphical illustration or need further details on any step?

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### Related Questions:
1. What is the general form for finding points of discontinuity in piecewise functions?
2. How does the behavior of logarithmic functions affect continuity in rational functions?
3. Why is the point \( x = 0 \) a non-removable discontinuity?
4. How do left-hand and right-hand limits help in determining continuity?
5. What are the different types of discontinuities a function can have?

### Tip:
When analyzing piecewise functions, always check the boundaries and individual definitions to determine continuity at each relevant point.

Найти точки разрыва функции f(x) = { x / lg(x + 1), x > -1; sin(πx), x ≤ -1 } и определить их характер.

Explore how to find and characterize the discontinuity points of a piecewise function with logarithmic and trigonometric components. This solution includes steps to analyze limits, identify non-removable discontinuities, and provides a graphical illustration.

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