\( \lim _ { x \rightarrow - \infty } \frac { 2 x + 2 n + x } { x - x ^ { 2 } } \)

We start with the limit \[ \lim_{x \rightarrow -\infty} \frac{2x + 2n + x}{x - x^2}. \] **Step 1: Simplify the numerator** Combine like terms in the numerator: \[ 2x + 2n + x = 3x + 2n. \] Thus, the limit becomes \[ \lim_{x \rightarrow -\infty} \frac{3x + 2n}{x - x^2}. \] **Step 2: Factor the denominator** Factor out \(x\) from the denominator: \[ x - x^2 = x(1 - x). \] The expression is now \[ \frac{3x + 2n}{x(1 - x)}. \] **Step 3: Separate the terms** We can split the expression into two fractions: \[ \frac{3x + 2n}{x(1 - x)} = \frac{3x}{x(1 - x)} + \frac{2n}{x(1 - x)} = \frac{3}{1 - x} + \frac{2n}{x(1 - x)}. \] **Step 4: Analyze the limit of each term** As \(x \rightarrow -\infty\): 1. For the first term, \(\frac{3}{1 - x}\): Since \(x\) is very large in magnitude (negative), \(1 - x\) tends to \(+\infty\). Therefore, \[ \frac{3}{1 - x} \rightarrow 0. \] 2. For the second term, \(\frac{2n}{x(1 - x)}\): Both \(x\) and \(1 - x\) grow in magnitude, making the denominator tend to \(\infty\). Hence, \[ \frac{2n}{x(1 - x)} \rightarrow 0. \] **Step 5: Combine the results** Since both terms approach 0, the overall limit is \[ \lim_{x \rightarrow -\infty} \frac{3x + 2n}{x - x^2} = 0. \] Thus, the final answer is \[ \boxed{0}. \]