mple: Show that f(x) is discontinuous at a = 2 where, f(x)= x^ 2 -x-2 x-2 & if x ne2\\ 1&x=2

We are given the piecewise function:

\[
f(x) = \begin{cases} 
\frac{x^2 - x - 2}{x - 2} & \text{if } x \neq 2 \\
1 & \text{if } x = 2 
\end{cases}
\]

We are tasked with showing that \( f(x) \) is discontinuous at \( a = 2 \).

### Step 1: Simplifying the function for \( x \neq 2 \)

The function for \( x \neq 2 \) is given as:

\[
f(x) = \frac{x^2 - x - 2}{x - 2}
\]

First, factor the quadratic expression in the numerator:

\[
x^2 - x - 2 = (x - 2)(x + 1)
\]

Thus, the function becomes:

\[
f(x) = \frac{(x - 2)(x + 1)}{x - 2}, \quad \text{for } x \neq 2
\]

Since \( x \neq 2 \), we can cancel the common factor \( x - 2 \):

\[
f(x) = x + 1, \quad \text{for } x \neq 2
\]

Thus, for \( x \neq 2 \), we have \( f(x) = x + 1 \).

### Step 2: Checking the limit of \( f(x) \) as \( x \to 2 \)

Next, we calculate the limit of \( f(x) \) as \( x \) approaches 2:

\[
\lim_{x \to 2} f(x) = \lim_{x \to 2} (x + 1) = 2 + 1 = 3
\]

### Step 3: Comparing the limit and the function value at \( x = 2 \)

We know that:

\[
f(2) = 1
\]

However, from Step 2, we found that:

\[
\lim_{x \to 2} f(x) = 3
\]

### Step 4: Conclusion of discontinuity

Since the limit of \( f(x) \) as \( x \to 2 \) is 3, but \( f(2) = 1 \), the function is not continuous at \( x = 2 \). Therefore, \( f(x) \) is **discontinuous at \( x = 2 \)**.

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Do you want more details on this process or have any questions?

### 5 Relative Questions:
1. What are the conditions for continuity of a function at a point?
2. How can we determine if a function is removable discontinuous at a point?
3. Can you explain the difference between a removable discontinuity and a jump discontinuity?
4. How would the function change if \( f(2) \) were redefined to match the limit?
5. Can you provide an example of a function that is continuous everywhere?

**Tip**: Always check both the limit and the actual function value to confirm if a function is continuous at a point.

Show that f(x) is discontinuous at a = 2 where, f(x)= x^2 - x - 2 / (x - 2) if x ≠ 2, and f(x) = 1 if x = 2.

Explore how to determine if a piecewise function f(x) = (x^2 - x - 2) / (x - 2) is discontinuous at x = 2. This problem delves into limits, factorization, and the definition of continuity. Suitable for Grades 10-12, the step-by-step solution guides students through identifying discontinuities using algebraic techniques and limit properties.

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