We are tasked with finding the limit:

$\lim_{x \to \infty} \frac{3x - 5}{\sqrt{2x^2 + 4x - 7}}$

Step 1: Analyzing the dominant terms

When $x \to \infty$, the highest powers of $x$ dominate both in the numerator and the denominator. In this case, the dominant term in the numerator is $3x$, and in the denominator, it's $\sqrt{2x^2} = \sqrt{2}x$.

Step 2: Simplifying the expression

To simplify the fraction, divide both the numerator and the denominator by $x$:

$\frac{3x - 5}{\sqrt{2x^2 + 4x - 7}} = \frac{x(3 - \frac{5}{x})}{x\sqrt{2 + \frac{4}{x} - \frac{7}{x^2}}}$

Now, factor out $x$ from the square root in the denominator:

$\frac{x(3 - \frac{5}{x})}{x\sqrt{2 + \frac{4}{x} - \frac{7}{x^2}}} = \frac{3 - \frac{5}{x}}{\sqrt{2 + \frac{4}{x} - \frac{7}{x^2}}}$

Step 3: Taking the limit as $x \to \infty$

As $x \to \infty$, the terms $\frac{5}{x}$, $\frac{4}{x}$, and $\frac{7}{x^2}$ all approach zero. Thus, the expression simplifies to:

$\frac{3}{\sqrt{2}}$

Final Answer:

$\lim_{x \to \infty} \frac{3x - 5}{\sqrt{2x^2 + 4x - 7}} = \frac{3}{\sqrt{2}}$

Let me know if you would like further details on this or have any other questions!

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Here are 5 related questions to extend your understanding:

1. How would the result change if the numerator were $5x - 3$?
2. What is the effect of changing the coefficient of $x^2$ in the denominator?
3. How do we handle limits at infinity when the numerator grows faster than the denominator?
4. Can L'Hopital's Rule be applied to this limit? Why or why not?
5. What happens if we replace $x \to \infty$ with $x \to -\infty$?

Tip: When taking limits at infinity, always focus on the highest powers of $x$ in both the numerator and denominator to identify the dominant behavior.