`lim_(x->oo)x^(3/2)(sqrt(x^3+1)-sqrt(x^3-1))`

Options

A. 1
B. `-1`
C. 0
D. none of these

Answer

To solve the limit \( \lim_{x \to \infty} x^{3/2} \left( \sqrt{x^3 + 1} - \sqrt{x^3 - 1} \right) \), we will follow these steps:

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

### Step 2: Multiply and divide by the conjugate
To simplify the expression, we multiply and divide by the conjugate:
\[ \lim_{x \to \infty} x^{3/2} \cdot \frac{\left( \sqrt{x^3 + 1} - \sqrt{x^3 - 1} \right) \left( \sqrt{x^3 + 1} + \sqrt{x^3 - 1} \right)}{\sqrt{x^3 + 1} + \sqrt{x^3 - 1}} \]

### Step 3: Simplify the numerator
Using the difference of squares, the numerator becomes:
\[ \sqrt{x^3 + 1}^2 - \sqrt{x^3 - 1}^2 = (x^3 + 1) - (x^3 - 1) = 2 \]
Thus, we rewrite the limit:
\[ \lim_{x \to \infty} \frac{2 x^{3/2}}{\sqrt{x^3 + 1} + \sqrt{x^3 - 1}} \]

### Step 4: Simplify the denominator
Now, we analyze the denominator:
\[ \sqrt{x^3 + 1} + \sqrt{x^3 - 1} \]
As \( x \to \infty \), both square roots can be approximated:
\[ \sqrt{x^3 + 1} \approx \sqrt{x^3} = x^{3/2} \quad \text{and} \quad \sqrt{x^3 - 1} \approx \sqrt{x^3} = x^{3/2} \]
Thus, the denominator becomes:
\[ \sqrt{x^3 + 1} + \sqrt{x^3 - 1} \approx x^{3/2} + x^{3/2} = 2x^{3/2} \]

### Step 5: Substitute back into the limit
Now we can substitute this back into our limit:
\[ \lim_{x \to \infty} \frac{2 x^{3/2}}{2 x^{3/2}} = \lim_{x \to \infty} 1 = 1 \]

### Conclusion
Thus, the limit is:
\[ \boxed{1} \]