Evaluate the following limits :
`lim_(x to 0)(sqrt(1+3x)-sqrt(1-3x))/(x)`

Answer

To evaluate the limit

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

we first check the form of the limit by substituting \(x = 0\):

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

Since we have an indeterminate form \(0/0\), we can apply L'Hôpital's Rule, which states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) results in \(0/0\) or \(\infty/\infty\), then:

\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \]

provided the limit on the right exists.

### Step 1: Differentiate the numerator and denominator

Let \(f(x) = \sqrt{1 + 3x} - \sqrt{1 - 3x}\) and \(g(x) = x\).

Now we find the derivatives \(f'(x)\) and \(g'(x)\):

1. **Differentiate \(f(x)\)**:
- Using the chain rule, the derivative of \(\sqrt{u}\) is \(\frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}\).
- For \(\sqrt{1 + 3x}\):
\[ f_1'(x) = \frac{1}{2\sqrt{1 + 3x}} \cdot 3 = \frac{3}{2\sqrt{1 + 3x}}. \]
- For \(\sqrt{1 - 3x}\):
\[ f_2'(x) = \frac{1}{2\sqrt{1 - 3x}} \cdot (-3) = -\frac{3}{2\sqrt{1 - 3x}}. \]
- Therefore,
\[ f'(x) = \frac{3}{2\sqrt{1 + 3x}} + \frac{3}{2\sqrt{1 - 3x}}. \]

2. **Differentiate \(g(x)\)**:
\[ g'(x) = 1. \]

### Step 2: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \left( \frac{3}{2\sqrt{1 + 3x}} + \frac{3}{2\sqrt{1 - 3x}} \right). \]

### Step 3: Evaluate the limit

Substituting \(x = 0\):

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

### Final Answer

Thus, the limit is

\[ \boxed{3}. \]