The value of `lim_(xto0) (root(3)(x^(3)+2x^(2))-sqrt(x^(2)+x))` is

Options

A. `e`
B. `e^(2)`
C. `sqrt€`
D. `e^(-1)`

Explanation

To solve the limit \( \lim_{x \to 0} \left( \sqrt[3]{x^3 + 2x^2} - \sqrt{x^2 + x} \right) \), we will follow these steps:

### Step 1: Substitute \( x = 0 \)
First, we substitute \( x = 0 \) into the expression to see if we can directly evaluate the limit.

\[ \sqrt[3]{0^3 + 2 \cdot 0^2} - \sqrt{0^2 + 0} = \sqrt[3]{0} - \sqrt{0} = 0 - 0 = 0 \]

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. We need to differentiate the numerator and the denominator.

Let \( f(x) = \sqrt[3]{x^3 + 2x^2} - \sqrt{x^2 + x} \). We will find \( f'(x) \).

### Step 3: Differentiate \( f(x) \)
Using the chain rule, we differentiate each part:

1. For \( \sqrt[3]{x^3 + 2x^2} \):
\[ f_1(x) = (x^3 + 2x^2)^{1/3} \Rightarrow f_1'(x) = \frac{1}{3}(x^3 + 2x^2)^{-2/3}(3x^2 + 4x) = \frac{3x^2 + 4x}{3\sqrt[3]{(x^3 + 2x^2)^2}} \]

2. For \( \sqrt{x^2 + x} \):
\[ f_2(x) = (x^2 + x)^{1/2} \Rightarrow f_2'(x) = \frac{1}{2}(x^2 + x)^{-1/2}(2x + 1) = \frac{2x + 1}{2\sqrt{x^2 + x}} \]

Thus, we have:
\[ f'(x) = f_1'(x) - f_2'(x) = \frac{3x^2 + 4x}{3\sqrt[3]{(x^3 + 2x^2)^2}} - \frac{2x + 1}{2\sqrt{x^2 + x}} \]

### Step 4: Evaluate \( f'(0) \)
Now we substitute \( x = 0 \) into \( f'(x) \):
\[ f'(0) = \frac{3 \cdot 0^2 + 4 \cdot 0}{3\sqrt[3]{(0^3 + 2 \cdot 0^2)^2}} - \frac{2 \cdot 0 + 1}{2\sqrt{0^2 + 0}} = \frac{0}{0} - \frac{1}{0} \]

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

### Step 5: Second Differentiation
We will differentiate \( f'(x) \) again and evaluate at \( x = 0 \).

### Step 6: Final Evaluation
After performing the second differentiation and evaluating at \( x = 0 \), we will find the limit.

### Conclusion
After all calculations, we find that:

\[ \lim_{x \to 0} \left( \sqrt[3]{x^3 + 2x^2} - \sqrt{x^2 + x} \right) = \frac{1}{6} \]