Check the continuity of `f(x) = {{:((2x^(2)-3x-2)/(x-2), if x ne 2), (5, if x = 2):}` at `x = 2`.

Explanation

To check the continuity of the function \( f(x) \) at \( x = 2 \), we need to verify three conditions:

1. The left-hand limit as \( x \) approaches 2.
2. The right-hand limit as \( x \) approaches 2.
3. The value of the function at \( x = 2 \).

The function is defined as:
\[ f(x) = \begin{cases} \frac{2x^2 - 3x - 2}{x - 2} & \text{if } x \neq 2 \\ 5 & \text{if } x = 2 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit
We need to find:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2x^2 - 3x - 2}{x - 2} \]

Substituting \( x = 2 - h \) (where \( h \to 0^+ \)):
\[ \lim_{h \to 0} \frac{2(2 - h)^2 - 3(2 - h) - 2}{(2 - h) - 2} \]

Expanding the numerator:
\[ = \lim_{h \to 0} \frac{2(4 - 4h + h^2) - 6 + 3h - 2}{-h} \]
\[ = \lim_{h \to 0} \frac{8 - 8h + 2h^2 - 6 + 3h - 2}{-h} \]
\[ = \lim_{h \to 0} \frac{2h^2 - 5h}{-h} \]
\[ = \lim_{h \to 0} - (2h - 5) = 5 \]

### Step 2: Calculate the Right-Hand Limit
Now we find:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{2x^2 - 3x - 2}{x - 2} \]

Substituting \( x = 2 + h \) (where \( h \to 0^+ \)):
\[ \lim_{h \to 0} \frac{2(2 + h)^2 - 3(2 + h) - 2}{(2 + h) - 2} \]

Expanding the numerator:
\[ = \lim_{h \to 0} \frac{2(4 + 4h + h^2) - 6 - 3h - 2}{h} \]
\[ = \lim_{h \to 0} \frac{8 + 8h + 2h^2 - 6 - 3h - 2}{h} \]
\[ = \lim_{h \to 0} \frac{2h^2 + 5h}{h} \]
\[ = \lim_{h \to 0} (2h + 5) = 5 \]

### Step 3: Value of the Function at \( x = 2 \)
From the definition of the function:
\[ f(2) = 5 \]

### Conclusion
Since:
\[ \lim_{x \to 2^-} f(x) = 5, \quad \lim_{x \to 2^+} f(x) = 5, \quad \text{and} \quad f(2) = 5 \]
All three values are equal, thus \( f(x) \) is continuous at \( x = 2 \).

### Final Answer
The function \( f(x) \) is continuous at \( x = 2 \).

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