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

Explanation

To determine whether the function \( f(x) \) is continuous at \( x = 2 \), we need to check the following:

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: Find the left-hand limit as \( x \) approaches 2.
We calculate the left-hand limit:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2x^2 - 3x - 2}{x - 2} \]

### Step 2: Substitute \( x = 2 - h \) (where \( h \to 0 \)).
This gives:
\[ \lim_{h \to 0} f(2 - h) = \lim_{h \to 0} \frac{2(2 - h)^2 - 3(2 - h) - 2}{(2 - h) - 2} \]

### Step 3: Simplify the expression.
Calculating \( (2 - h)^2 \):
\[ (2 - h)^2 = 4 - 4h + h^2 \]
Substituting this back into the limit:
\[ = \lim_{h \to 0} \frac{2(4 - 4h + h^2) - 3(2 - h) - 2}{-h} \]
\[ = \lim_{h \to 0} \frac{8 - 8h + 2h^2 - 6 + 3h - 2}{-h} \]
\[ = \lim_{h \to 0} \frac{2h^2 - 5h}{-h} \]

### Step 4: Factor out \( h \) from the numerator.
\[ = \lim_{h \to 0} \frac{h(2h - 5)}{-h} \]
Cancelling \( h \) (as \( h \neq 0 \)):
\[ = \lim_{h \to 0} -(2h - 5) = -(-5) = 5 \]

### Step 5: Find the value of the function at \( x = 2 \).
From the definition of the function:
\[ f(2) = 5 \]

### Step 6: Compare the left-hand limit and the value of the function.
Since:
\[ \lim_{x \to 2^-} f(x) = 5 \quad \text{and} \quad f(2) = 5 \]
Both limits are equal.

### Conclusion:
The function \( f(x) \) is continuous at \( x = 2 \).

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