Works 2. (Yellow Paper) 1. Discuss the continuity of $$ f $$ at $$ l a) x=-3,(b) \mid x=0 $$ and $$ |c| x=1 $$ it discontinuous, give the type of discontinuity. 5 pointseach Let f be the function defines belbm. \[ f(x)=\left\{\begin{array}{ccc} \left|\frac{x^{2}+3 x}{x-3}\right| & \text { if } & x \leq 0, x \neq-3 \ x+1 & \text { if } & 0

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Let's analyze the continuity of the function $$ f(x) $$ at the points $$ x = -3 $$, $$ x = 0 $$, and $$ x = 1 $$. The function is defined as:

$$
f(x)=\left\{\begin{array}{ccc}
\left|\frac{x^{2}+3x}{x-3}\right| & \text{if} & x \leq 0, \, x \neq -3 \
x + 1 & \text{if} & 0 < x < 1 \
\sqrt{x} & \text{if} & x \geq 1
\end{array}\right.
$$

### (a) Continuity at $$ x = -3 $$

**Function Behavior:**
- For $$ x \leq 0 $$ and $$ x \neq -3 $$:
  $$
  f(x) = \left|\frac{x^{2}+3x}{x-3}\right|
  $$
- At $$ x = -3 $$:
  - The function $$ f(x) $$ is **not defined** because $$ x = -3 $$ is excluded from this piece.

**Limits:**
$$
\lim_{{x \to -3^-}} f(x) = \lim_{{x \to -3}} \left|\frac{(-3)^2 + 3(-3)}{-3 - 3}\right| = \left|\frac{9 - 9}{-6}\right| = 0
$$

**Conclusion:**
- **Discontinuity Type:** Removable Discontinuity
- **Reason:** The limit exists as $$ x $$ approaches $$-3$$ (which is 0), but $$ f(-3) $$ is not defined. If we define $$ f(-3) = 0 $$, the function becomes continuous at this point.

### (b) Continuity at $$ x = 0 $$

**Function Behavior:**
- For $$ x \leq 0 $$:
  $$
  f(0) = \left|\frac{0^2 + 3(0)}{0 - 3}\right| = 0
  $$
- For $$ 0 < x < 1 $$:
  $$
  \lim_{{x \to 0^+}} f(x) = 0 + 1 = 1
  $$

**Limits:**
$$
\lim_{{x \to 0^-}} f(x) = 0 \quad \text{and} \quad \lim_{{x \to 0^+}} f(x) = 1
$$

**Conclusion:**
- **Discontinuity Type:** Jump Discontinuity
- **Reason:** The left-hand limit and right-hand limit at $$ x = 0 $$ are not equal. The function "jumps" from 0 to 1.

### (c) Continuity at $$ x = 1 $$

**Function Behavior:**
- For $$ 0 < x < 1 $$:
  $$
  \lim_{{x \to 1^-}} f(x) = 1 + 1 = 2
  $$
- For $$ x \geq 1 $$:
  $$
  f(1) = \sqrt{1} = 1
  $$

**Limits:**
$$
\lim_{{x \to 1^-}} f(x) = 2 \quad \text{and} \quad \lim_{{x \to 1^+}} f(x) = 1
$$

**Conclusion:**
- **Discontinuity Type:** Jump Discontinuity
- **Reason:** The left-hand limit and right-hand limit at $$ x = 1 $$ are not equal. The function "jumps" from 2 to 1.

---

### Summary

1. **At $$ x = -3 $$:** Removable Discontinuity
2. **At $$ x = 0 $$:** Jump Discontinuity
3. **At $$ x = 1 $$:** Jump Discontinuity
At $$ x = -3 $$, there's a removable discontinuity. At $$ x = 0 $$ and $$ x = 1 $$, there are jump discontinuities.

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