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.

Explore the detailed step-by-step process of computing the limit lim x->-inf sqrt(9x^2 + 5x) - 3x. Learn about simplifying square roots, factoring, and evaluating limits for large negative values of x. This problem is suitable for advanced high school students in Grades 11-12.

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