`lim_(x to 0) (root(3)(1+sin x)- root(3)(1-sinx))/x`=

Options

A. `1/3`
B. `2/3`
C. 1
D. 0

Answer

To solve the limit

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

we can follow these steps:

### Step 1: Identify the form of the limit
When we substitute \(x = 0\) directly into the expression, we get:

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

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

1. Differentiate the numerator:
\[ \frac{d}{dx} \left( \sqrt[3]{1 + \sin x} - \sqrt[3]{1 - \sin x} \right) = \frac{1}{3}(1 + \sin x)^{-2/3} \cos x - \frac{1}{3}(1 - \sin x)^{-2/3} (-\cos x) \]

Simplifying this gives:
\[ \frac{1}{3} \left( (1 + \sin x)^{-2/3} \cos x + (1 - \sin x)^{-2/3} \cos x \right) \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(x) = 1 \]

Now we can rewrite our limit using these derivatives:

\[ \lim_{x \to 0} \frac{\frac{1}{3} \left( (1 + \sin x)^{-2/3} \cos x + (1 - \sin x)^{-2/3} \cos x \right)}{1} \]

### Step 3: Evaluate the limit
Now we substitute \(x = 0\) into the new expression:

- At \(x = 0\), \(\sin(0) = 0\) and \(\cos(0) = 1\):
\[ \lim_{x \to 0} \frac{1}{3} \left( (1 + 0)^{-2/3} \cdot 1 + (1 - 0)^{-2/3} \cdot 1 \right) = \frac{1}{3} \left( 1 + 1 \right) = \frac{1}{3} \cdot 2 = \frac{2}{3} \]

### Final Answer
Thus, the limit is:

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