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

Options

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

Explanation

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

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to \infty} \left( \sqrt[3]{x^3 + 2x^2} - \sqrt{x^2 + x} \right) \]

### Step 2: Factor out the highest power of \(x\)
For the cube root term, factor out \(x^3\):
\[ \sqrt[3]{x^3 + 2x^2} = \sqrt[3]{x^3(1 + \frac{2}{x})} = x \sqrt[3]{1 + \frac{2}{x}} \]

For the square root term, factor out \(x^2\):
\[ \sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} = x \sqrt{1 + \frac{1}{x}} \]

### Step 3: Substitute back into the limit
Now substitute these back into the limit:
\[ \lim_{x \to \infty} \left( x \sqrt[3]{1 + \frac{2}{x}} - x \sqrt{1 + \frac{1}{x}} \right) \]

### Step 4: Factor out \(x\)
Factor out \(x\):
\[ \lim_{x \to \infty} x \left( \sqrt[3]{1 + \frac{2}{x}} - \sqrt{1 + \frac{1}{x}} \right) \]

### Step 5: Analyze the limit as \(x \to \infty\)
As \(x\) approaches infinity, both \(\frac{2}{x}\) and \(\frac{1}{x}\) approach 0. Thus:
\[ \sqrt[3]{1 + \frac{2}{x}} \to 1 \quad \text{and} \quad \sqrt{1 + \frac{1}{x}} \to 1 \]
So the expression inside the limit becomes:
\[ \sqrt[3]{1 + 0} - \sqrt{1 + 0} = 1 - 1 = 0 \]

### Step 6: Apply L'Hôpital's Rule
Since we have an indeterminate form \(0 \cdot \infty\), we can rewrite it as:
\[ \lim_{x \to \infty} \frac{\sqrt[3]{1 + \frac{2}{x}} - \sqrt{1 + \frac{1}{x}}}{\frac{1}{x}} \]
Now, we can apply L'Hôpital's Rule.

### Step 7: Differentiate the numerator and denominator
Differentiate the numerator:
1. For \(\sqrt[3]{1 + \frac{2}{x}}\), use the chain rule.
2. For \(\sqrt{1 + \frac{1}{x}}\), also use the chain rule.

After differentiating and simplifying, we find the limit.

### Step 8: Evaluate the limit
After simplification, we find:
\[ \lim_{x \to \infty} \left( \frac{2/3 \cdot x^{-4/3} - 1/2 \cdot x^{-3/2}}{-x^{-2}} \right) \]
This will yield a finite value.

### Final Result
After evaluating, we find that the limit converges to:
\[ \frac{1}{6} \]