`lim_(x to 0) (sqrt(1 + 3x) + sqrt(1 - 3x))/(1 + 3x)` is equal to

Options

A. 2
B. 1
C. 0
D. `-1`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sqrt{1 + 3x} + \sqrt{1 - 3x}}{1 + 3x} \), we will follow these steps:

### Step 1: Substitute \( x = 0 \) in the expression
We start by substituting \( x = 0 \) into the expression to see if we can directly evaluate the limit.

\[ \sqrt{1 + 3(0)} + \sqrt{1 - 3(0)} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2 \]

The denominator becomes:

\[ 1 + 3(0) = 1 \]

So, we have:

\[ \frac{2}{1} = 2 \]

### Step 2: Conclusion
Since we can directly substitute \( x = 0 \) into the limit without encountering any indeterminate forms, we find that:

\[ \lim_{x \to 0} \frac{\sqrt{1 + 3x} + \sqrt{1 - 3x}}{1 + 3x} = 2 \]

Thus, the final answer is:

\[ \boxed{2} \]