The value of `lim_(xrarr oo) (3sqrt(x^3+x^2)-3sqrt(x^3-x^2))`, is

Options

A. `(1)/(2)`
B. `(2)/(3)`
C. `1`
D. `(4)/(3)`

Explanation

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

### Step 1: Factor out the common terms
We can factor out \( 3 \) from the expression:
\[ \lim_{x \to \infty} 3(\sqrt{x^3+x^2} - \sqrt{x^3-x^2}) \]

### Step 2: Simplify the square roots
Next, we simplify the square roots inside the limit:
\[ \sqrt{x^3+x^2} = \sqrt{x^3(1+\frac{1}{x})} = x^{3/2}\sqrt{1+\frac{1}{x}} \]
\[ \sqrt{x^3-x^2} = \sqrt{x^3(1-\frac{1}{x})} = x^{3/2}\sqrt{1-\frac{1}{x}} \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to \infty} 3\left(x^{3/2}\sqrt{1+\frac{1}{x}} - x^{3/2}\sqrt{1-\frac{1}{x}}\right) \]

### Step 3: Factor out \( x^{3/2} \)
Now, we can factor \( x^{3/2} \) out of the expression:
\[ \lim_{x \to \infty} 3x^{3/2}\left(\sqrt{1+\frac{1}{x}} - \sqrt{1-\frac{1}{x}}\right) \]

### Step 4: Analyze the limit of the square root difference
As \( x \to \infty \), \( \frac{1}{x} \to 0 \). Therefore, we can use the binomial expansion for the square roots:
\[ \sqrt{1+\frac{1}{x}} \approx 1 + \frac{1}{2x} \quad \text{and} \quad \sqrt{1-\frac{1}{x}} \approx 1 - \frac{1}{2x} \]
Thus,
\[ \sqrt{1+\frac{1}{x}} - \sqrt{1-\frac{1}{x}} \approx \left(1 + \frac{1}{2x}\right) - \left(1 - \frac{1}{2x}\right) = \frac{1}{x} \]

### Step 5: Substitute back into the limit
Substituting back into the limit gives:
\[ \lim_{x \to \infty} 3x^{3/2}\left(\frac{1}{x}\right) = \lim_{x \to \infty} 3x^{1/2} = 3\sqrt{x} \]

### Step 6: Evaluate the limit as \( x \to \infty \)
As \( x \to \infty \), \( 3\sqrt{x} \to \infty \).

### Final Result
Thus, the value of the limit is:
\[ \lim_{x \to \infty} (3\sqrt{x^3+x^2} - 3\sqrt{x^3-x^2}) = \infty \]