Find the limit: \( \lim _{x \rightarrow \infty}\left(\sqrt{x^{2}+5 x}-\sqrt{x^{2}-5 x}\right) \)

To evaluate the limit \[ L = \lim_{x \rightarrow \infty}\left(\sqrt{x^{2}+5x}-\sqrt{x^{2}-5x}\right), \] we can start by simplifying the expression under the limit. A common strategy for limits involving square roots is to multiply by the conjugate. We multiply the expression by \[ \frac{\sqrt{x^2 + 5x} + \sqrt{x^2 - 5x}}{\sqrt{x^2 + 5x} + \sqrt{x^2 - 5x}}, \] to rationalize the difference of the square roots: \[ L = \lim_{x \rightarrow \infty} \frac{\left(\sqrt{x^2 + 5x} - \sqrt{x^2 - 5x}\right)\left(\sqrt{x^2 + 5x} + \sqrt{x^2 - 5x}\right)}{\sqrt{x^2 + 5x} + \sqrt{x^2 - 5x}}. \] The numerator simplifies to \[ (\sqrt{x^2 + 5x})^2 - (\sqrt{x^2 - 5x})^2 = (x^2 + 5x) - (x^2 - 5x) = 10x. \] Thus, we have: \[ L = \lim_{x \rightarrow \infty} \frac{10x}{\sqrt{x^2 + 5x} + \sqrt{x^2 - 5x}}. \] Next, we can simplify the denominator: \[ \sqrt{x^2 + 5x} = \sqrt{x^2(1 + \frac{5}{x})} = x\sqrt{1 + \frac{5}{x}}, \] and \[ \sqrt{x^2 - 5x} = \sqrt{x^2(1 - \frac{5}{x})} = x\sqrt{1 - \frac{5}{x}}. \] Now, substituting back into our expression for \(L\): \[ L = \lim_{x \rightarrow \infty} \frac{10x}{x\left(\sqrt{1 + \frac{5}{x}} + \sqrt{1 - \frac{5}{x}}\right)} = \lim_{x \rightarrow \infty} \frac{10}{\sqrt{1 + \frac{5}{x}} + \sqrt{1 - \frac{5}{x}}}. \] As \(x \rightarrow \infty\), both \(\frac{5}{x} \rightarrow 0\), so the limit becomes: \[ L = \frac{10}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{10}{1 + 1} = \frac{10}{2} = 5. \] Thus, the final answer is: \[ \boxed{5}. \]