The given expression is:

$\lim_{{x \to \infty}} \frac{\sqrt{x + 2x^2}}{7x - 1}$

Let's solve this step by step.

Step 1: Simplify the expression inside the square root.

We have $\sqrt{x + 2x^2}$. Factor out $x^2$ from the expression inside the square root:

$\sqrt{x + 2x^2} = \sqrt{x(1 + 2x)} = x\sqrt{1 + \frac{1}{2x}}$

As $x \to \infty$, $\frac{1}{2x}$ approaches 0, so:

$\sqrt{1 + \frac{1}{2x}} \to \sqrt{1} = 1$

Thus, for large $x$, the square root simplifies to:

$\sqrt{x + 2x^2} \approx x$

Step 2: Simplify the denominator.

In the denominator, $7x - 1$, as $x \to \infty$, the constant $-1$ becomes negligible compared to $7x$. So:

$7x - 1 \approx 7x$

Step 3: Simplify the whole expression.

Now, substituting these approximations into the original limit expression:

$\lim_{{x \to \infty}} \frac{x}{7x} = \lim_{{x \to \infty}} \frac{1}{7} = \frac{1}{7}$

Final Answer:

The limit is:

$\boxed{\frac{1}{7}}$

Would you like more details or explanations on any specific step?

Here are five related questions:

1. What happens if the degrees of the numerator and denominator in a limit are not the same?
2. How would the limit change if the denominator were $7x + 1$ instead of $7x - 1$?
3. Can you explain the behavior of limits at infinity for rational functions?
4. How does L'Hopital's Rule help in solving indeterminate limits?
5. How would this limit be affected if the expression inside the square root had a different leading term?

Tip: When simplifying limits at infinity, factor out the highest power of $x$ from both the numerator and the denominator to help see which terms dominate as $x \to \infty$.