Evaluate \( \lim _{x \rightarrow \infty}\left(\sqrt{x^{4}+x^{3}}-\sqrt{x^{4}-x^{3}}\right) \)

To evaluate the limit \[ \lim _{x \rightarrow \infty}\left(\sqrt{x^{4}+x^{3}}-\sqrt{x^{4}-x^{3}}\right), \] we start by simplifying the expression inside the limit. First, we will factor out \( x^4 \) from both square roots: \[ \sqrt{x^{4}+x^{3}} = \sqrt{x^4(1 + \frac{1}{x})} = x^2\sqrt{1 + \frac{1}{x}} \] and \[ \sqrt{x^{4}-x^{3}} = \sqrt{x^4(1 - \frac{1}{x})} = x^2\sqrt{1 - \frac{1}{x}}. \] Now we can rewrite the expression: \[ \sqrt{x^{4}+x^{3}} - \sqrt{x^{4}-x^{3}} = x^2\sqrt{1 + \frac{1}{x}} - x^2\sqrt{1 - \frac{1}{x}}. \] Factor out \( x^2 \): \[ = x^2 \left( \sqrt{1 + \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} \right). \] Next, we need to evaluate the limit as \( x \to \infty \): \[ \lim_{x \to \infty} x^2 \left( \sqrt{1 + \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} \right). \] To proceed, we can use the fact that for small \( y \), \( \sqrt{1 + y} \approx 1 + \frac{1}{2}y \) and \( \sqrt{1 - y} \approx 1 - \frac{1}{2}y \). Letting \( y = \frac{1}{x} \), we have: \[ \sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2} \cdot \frac{1}{x} \quad \text{and} \quad \sqrt{1 - \frac{1}{x}} \approx 1 - \frac{1}{2} \cdot \frac{1}{x}. \] Thus, \[ \sqrt{1 + \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} \approx \left(1 + \frac{1}{2x}\right) - \left(1 - \frac{1}{2x}\right) = \frac{1}{2x} + \frac{1}{2x} = \frac{1}{x}. \] Now substituting back into our limit: \[ \lim_{x \to \infty} x^2 \left( \sqrt{1 + \frac{1}{x}} - \sqrt{1 - \frac{1}{x}} \right) \approx \lim_{x \to \infty} x^2 \cdot \frac{1}{x} = \lim_{x \to \infty} x = \infty. \] Thus, the limit evaluates to: \[ \boxed{\infty}. \]