If `f(x) = (x^(2)-1)/(x-1)` Discuss the continuity at `x rarr 1`

Answer

To discuss the continuity of the function \( f(x) = \frac{x^2 - 1}{x - 1} \) at \( x \to 1 \), we need to check the left-hand limit, right-hand limit, and the value of the function at that point.

### Step 1: Evaluate the left-hand limit as \( x \to 1^- \)

The left-hand limit is given by:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x^2 - 1}{x - 1} \]
Since \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \), we can rewrite the function:
\[ f(x) = \frac{(x - 1)(x + 1)}{x - 1} \]
For \( x \neq 1 \), this simplifies to:
\[ f(x) = x + 1 \]
Thus, we can find the left-hand limit:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x + 1) = 1 + 1 = 2 \]

### Step 2: Evaluate the right-hand limit as \( x \to 1^+ \)

The right-hand limit is given by:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x^2 - 1}{x - 1} \]
Using the same simplification as before:
\[ f(x) = x + 1 \]
Thus, we can find the right-hand limit:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x + 1) = 1 + 1 = 2 \]

### Step 3: Evaluate the function value at \( x = 1 \)

Now we need to find \( f(1) \):
\[ f(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0} \]
This is an indeterminate form, so we cannot directly evaluate \( f(1) \). However, we can say that \( f(1) \) is not defined.

### Step 4: Conclusion on continuity

For \( f(x) \) to be continuous at \( x = 1 \), the following must hold:
1. \( \lim_{x \to 1^-} f(x) = 2 \)
2. \( \lim_{x \to 1^+} f(x) = 2 \)
3. \( f(1) \) must exist and be equal to these limits.

Since \( f(1) \) is not defined, we conclude that \( f(x) \) is not continuous at \( x = 1 \).

### Summary

- Left-hand limit: \( 2 \)
- Right-hand limit: \( 2 \)
- Function value at \( x = 1 \): Not defined

Thus, the function \( f(x) \) is **not continuous** at \( x = 1 \).

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