\( \lim _ { x \rightarrow \infty } \ln ( \frac { x ^ { 2 } + 3 } { x ^ { 2 } + 4 x } ) ^ { \frac { x ^ { 2 } - 1 } { x } } \)

To evaluate the limit \[ \lim _ { x \rightarrow \infty } \ln \left( \frac { x ^ { 2 } + 3 } { x ^ { 2 } + 4 x } \right) ^ { \frac { x ^ { 2 } - 1 } { x } }, \] we can simplify the expression. First, we'll simplify the argument of the logarithm: \[ \frac { x^2 + 3 } { x^2 + 4x } = \frac { 1 + \frac{3}{x^2} } { 1 + \frac{4}{x} }. \] As \( x \) approaches infinity, both \(\frac{3}{x^2}\) and \(\frac{4}{x}\) approach 0. Therefore, \[ \frac { x^2 + 3 } { x^2 + 4x } \to \frac{1 + 0}{1 + 0} = 1. \] Since the limit of the fraction approaches 1, the logarithm of that fraction will approach 0: \[ \ln \left( \frac { x ^ { 2 } + 3 } { x ^ { 2 } + 4 x } \right) \to \ln(1) = 0. \] Now let's consider the exponent \(\frac{x^2 - 1}{x}\): \[ \frac{x^2 - 1}{x} = x - \frac{1}{x}. \] As \( x \) approaches infinity, \( x - \frac{1}{x} \) approaches \( x \). Putting this all together, we have: \[ \ln \left( \frac { x ^ { 2 } + 3 } { x ^ { 2 } + 4 x } \right)^{\frac{x^2 - 1}{x}} = \left( x - \frac{1}{x} \right) \ln \left(\frac { x ^ { 2 } + 3 } { x ^ { 2 } + 4 x }\right). \] The logarithm approaches 0, and when this is multiplied by \( x - \frac{1}{x} \) (which approaches \( x \)), we find ourselves looking at \( \infty \cdot 0 \), which is an indeterminate form. To resolve this, we can consider the limit: \[ = \lim_{x \to \infty} (x - \frac{1}{x}) \ln \left( \frac{1 + \frac{3}{x^2}}{1 + \frac{4}{x}} \right). \] Using the Taylor series expansion for \(\ln(1 + y) \approx y\) when \( y \) is small, we can approximate: \[ \ln \left( \frac{1 + \frac{3}{x^2}}{1 + \frac{4}{x}} \right) \approx \ln \left( 1 - \frac{4}{x} + O\left(\frac{1}{x^2}\right) \right) \approx -\frac{4}{x}. \] Thus we have: \[ = \lim_{x \to \infty} (x - \frac{1}{x}) \left(-\frac{4}{x}\right) \approx -4 \lim_{x \to \infty} \left(1 - \frac{1}{x^2}\right) = -4. \] Through all this simplification, our original limit resolves to: \[ \boxed{-4}. \]