Let's solve the given limit:

$\lim_{x \to +\infty} \sqrt{x^3} \left[ \frac{3}{\sqrt{x}} - \arctan \left( \frac{3}{\sqrt{x}} \right) \right]$

Step-by-step Solution:

1. Simplify the expression inside the limit:

   • We can rewrite $\sqrt{x^3}$ as $x^{3/2}$.
   • Thus, the expression becomes: $\lim_{x \to +\infty} x^{3/2} \left[ \frac{3}{\sqrt{x}} - \arctan \left( \frac{3}{\sqrt{x}} \right) \right]$

2. Simplify $\frac{3}{\sqrt{x}}$:

   • $\frac{3}{\sqrt{x}} = 3x^{-1/2}$, so now the limit becomes: $\lim_{x \to +\infty} x^{3/2} \left[ 3x^{-1/2} - \arctan \left( 3x^{-1/2} \right) \right]$

3. Evaluate the limit as $x \to +\infty$:

   • As $x \to +\infty$, $x^{-1/2} \to 0$.
   • Hence, $3x^{-1/2} \to 0$, and $\arctan(3x^{-1/2}) \to 0$ since $\arctan(y) \to 0$ as $y \to 0$.
   • Thus, the expression inside the square brackets approaches: $3x^{-1/2} - \arctan(3x^{-1/2}) \to 0 \text{ as } x \to +\infty.$
   • Now let's handle the outer factor $x^{3/2}$.

4. Analyze the asymptotic behavior:

   • For large $x$, the difference $3x^{-1/2} - \arctan(3x^{-1/2})$ behaves like $3x^{-1/2}$, since $\arctan(z) \approx z$ for small $z$.
   • Hence, for large $x$, the product $x^{3/2} \cdot 3x^{-1/2}$ behaves like $3x$.

5. Final limit:

   • Therefore, the limit as $x \to +\infty$ is: $\lim_{x \to +\infty} x^{3/2} \cdot 3x^{-1/2} = 3x.$
   • As $x \to +\infty$, this grows without bound. Therefore, the limit is:

$\boxed{\infty}.$

Would you like more details or clarifications?

Related Questions:

1. How does $\arctan(x)$ behave for very large and very small values of $x$?
2. What happens if we change the power of $x$ in $\sqrt{x^3}$ to another power?
3. How would the limit change if we used a different trigonometric function like $\sin^{-1}(x)$?
4. How do we apply L'Hopital's Rule in similar limit problems involving infinity?
5. What are the key asymptotic behaviors of functions involving $\arctan(x)$ and $x^{n/2}$?

Tip:

When dealing with limits at infinity involving complex functions like $\arctan$, approximations for small or large values can simplify the problem dramatically.