If `f(x) = (sqrt(x+3)-2)/(x^(3)-1) , x != 1`, is continuous at x = 1, then f(1) is

Options

A. 12
B. `1/8`
C. `1/12`
D. `8`

Answer

To determine the value of \( f(1) \) such that the function \( f(x) = \frac{\sqrt{x+3} - 2}{x^3 - 1} \) is continuous at \( x = 1 \), we need to find the limit of \( f(x) \) as \( x \) approaches 1.

### Step-by-Step Solution:

1. **Identify the limit**:
We need to find:
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{\sqrt{x+3} - 2}{x^3 - 1} \]

2. **Substituting \( x = 1 \)**:
First, let's check what happens when we substitute \( x = 1 \):
\[ f(1) = \frac{\sqrt{1+3} - 2}{1^3 - 1} = \frac{\sqrt{4} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0} \]
This is an indeterminate form, so we need to simplify the expression.

3. **Rationalizing the numerator**:
To eliminate the indeterminate form, we can rationalize the numerator:
\[ \frac{\sqrt{x+3} - 2}{x^3 - 1} \cdot \frac{\sqrt{x+3} + 2}{\sqrt{x+3} + 2} = \frac{(\sqrt{x+3} - 2)(\sqrt{x+3} + 2)}{(x^3 - 1)(\sqrt{x+3} + 2)} \]
The numerator simplifies to:
\[ (\sqrt{x+3})^2 - 2^2 = x + 3 - 4 = x - 1 \]
So we have:
\[ \lim_{x \to 1} \frac{x - 1}{(x^3 - 1)(\sqrt{x+3} + 2)} \]

4. **Factoring the denominator**:
The denominator \( x^3 - 1 \) can be factored as:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]
Therefore, we can rewrite the limit:
\[ \lim_{x \to 1} \frac{x - 1}{(x - 1)(x^2 + x + 1)(\sqrt{x+3} + 2)} \]
We can cancel \( x - 1 \) from the numerator and denominator:
\[ \lim_{x \to 1} \frac{1}{(x^2 + x + 1)(\sqrt{x+3} + 2)} \]

5. **Substituting \( x = 1 \) again**:
Now we can substitute \( x = 1 \):
\[ = \frac{1}{(1^2 + 1 + 1)(\sqrt{1+3} + 2)} = \frac{1}{(1 + 1 + 1)(2 + 2)} = \frac{1}{3 \cdot 4} = \frac{1}{12} \]

6. **Conclusion**:
Since the limit exists and is equal to \( f(1) \), we have:
\[ f(1) = \frac{1}{12} \]

### Final Answer:
Thus, \( f(1) \) is \( \frac{1}{12} \).