We need to compute the following limit:

$\lim_{{x \to -\infty}} \left( \sqrt{9x^2 + 5x} - 3x \right)$

Step-by-step solution:

1. Factor the expression inside the square root:

   First, we factor out $x^2$ from the square root term:

   $\sqrt{9x^2 + 5x} = \sqrt{x^2(9 + \frac{5}{x})} = |x|\sqrt{9 + \frac{5}{x}}$

   Since $x \to -\infty$, we know that $|x| = -x$ (because $x$ is negative). Thus, we get:

   $\sqrt{9x^2 + 5x} = -x\sqrt{9 + \frac{5}{x}}$

2. Rewrite the original limit expression:

   Substituting this into the original expression, we have:

   $\lim_{{x \to -\infty}} \left( -x\sqrt{9 + \frac{5}{x}} - 3x \right)$

   Factor out $-x$ from both terms:

   $\lim_{{x \to -\infty}} -x\left( \sqrt{9 + \frac{5}{x}} + 3 \right)$

3. Evaluate the limit:

   As $x \to -\infty$, $\frac{5}{x} \to 0$. So, $\sqrt{9 + \frac{5}{x}} \to \sqrt{9} = 3$. Substituting this in:

   $\lim_{{x \to -\infty}} -x(3 + 3) = \lim_{{x \to -\infty}} -x(6) = -6x$

   As $x \to -\infty$, $-6x \to +\infty$.

Thus, the limit is:

$\lim_{{x \to -\infty}} \left( \sqrt{9x^2 + 5x} - 3x \right) = \infty$

Would you like a more detailed explanation, or do you have any questions?

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Here are some related questions for practice:

1. Compute $\lim_{{x \to \infty}} \left( \sqrt{4x^2 + x} - 2x \right)$.
2. Evaluate $\lim_{{x \to -\infty}} \left( \sqrt{x^2 + 10x} - x \right)$.
3. Find the limit $\lim_{{x \to \infty}} \left( \sqrt{x^2 + 3x + 1} - x \right)$.
4. Compute $\lim_{{x \to -\infty}} \left( \sqrt{16x^2 + 7x} - 4x \right)$.
5. Evaluate $\lim_{{x \to \infty}} \left( \sqrt{25x^2 + 2x} - 5x \right)$.

Tip: When solving limits involving square roots and infinity, it often helps to factor out the highest power of $x$ and simplify the expression.