2.1. Define continuity of a function $$ f(x) $$ at point $$ x=a $$. 2.2. Let $$ f(x)=\left\{\begin{array}{ll}-1 & ; x \leq 0 \ x^{2} & ; 0

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**2.1. Definition of Continuity**

A function $$ f(x) $$ is said to be continuous at a point $$ x=a $$ if the following three conditions are satisfied:
1. $$ f(a) $$ is defined.
2. The limit $$ \lim_{x \to a} f(x) $$ exists.
3. $$ \lim_{x \to a} f(x) = f(a) $$.

---

**2.2. Analysis of $$ f(x) $$**

The function is given by
$$
f(x)=
\begin{cases}
-1 & ;\ x \leq 0, \
x^{2} & ;\ 02.
\end{cases}
$$

**2.2.1. At $$ x=1 $$:**

- **Determine $$ f(1) $$:**  
  Since $$ 1 $$ satisfies $$ 1 \leq x < 2 $$, we use $$ f(x)=x $$. Thus,
  $$
  f(1)=1.
  $$

- **Left-hand limit as $$ x\to 1^- $$:**  
  For $$ 0 < x < 1 $$, $$ f(x)=x^{2} $$. Therefore,
  $$
  \lim_{x \to 1^-} f(x)=\lim_{x \to 1^-} x^2=1.
  $$

- **Right-hand limit as $$ x\to 1^+ $$:**  
  For $$ 1 \leq x < 2 $$, $$ f(x)=x $$. Thus,
  $$
  \lim_{x \to 1^+} f(x)=\lim_{x \to 1^+} x=1.
  $$

Since both one-sided limits exist and are equal, and $$ \lim_{x \to 1} f(x)= f(1)=1 $$, the function is continuous at $$ x=1 $$.

---

**2.2.2. At $$ x=2 $$:**

- **Determine $$ f(2) $$:**  
  The definitions provide $$ f(x)=x $$ for $$ 1 \leq x<2 $$ and $$ f(x)=4-x $$ for $$ x>2 $$. Notice that $$ x=2 $$ does not fall into either interval. Hence, $$ f(2) $$ is not defined.

- **Check the limits around $$ x=2 $$:**

- For $$ x \to 2^- $$ (using $$ f(x)=x $$):
    $$
    \lim_{x\to 2^-} f(x)=2.
    $$
  
  - For $$ x \to 2^+ $$ (using $$ f(x)=4-x $$):
    $$
    \lim_{x\to 2^+} f(x)=4-2=2.
    $$

The two-sided limit exists and equals $$ 2 $$. However, since $$ f(2) $$ is not defined, the function has a removable discontinuity at $$ x=2 $$.

---

**2.3. Analysis of $$ g(x) $$ in $$[2,4]$$**

The function is defined by
$$
g(x)=
\begin{cases}
1 & ;\ x \leq -2, \
\frac{x}{2} & ;\ -2**2.1. Continuity of a Function at a Point** A function $$ f(x) $$ is continuous at a point $$ x=a $$ if: 1. $$ f(a) $$ is defined. 2. The limit $$ \lim_{x \to a} f(x) $$ exists. 3. $$ \lim_{x \to a} f(x) = f(a) $$. --- **2.2. Continuity of $$ f(x) $$ at Specific Points** Given: \[ f(x)= \begin{cases} -1 & ;\ x \leq 0, \ x^{2} & ;\ 02. \end{cases} $$ **2.2.1. At $$ x=1 $$:** - $$ f(1)=1 $$. - $$ \lim_{x \to 1^-} f(x)=1 $$. - $$ \lim_{x \to 1^+} f(x)=1 $$. - Since all conditions are met, $$ f(x) $$ is **continuous** at $$ x=1 $$. **2.2.2. At $$ x=2 $$:** - $$ f(2) $$ is **undefined**. - $$ \lim_{x \to 2^-} f(x)=2 $$. - $$ \lim_{x \to 2^+} f(x)=2 $$. - The function has a **removable discontinuity** at $$ x=2 $$ because $$ f(2) $$ is not defined, but the limits exist and are equal. --- **2.3. Continuity of $$ g(x) $$ on $$[2,4]$$** Given: \[ g(x)= \begin{cases} 1 & ;\ x \leq -2, \ \frac{x}{2} & ;\ -2

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