To solve the limit $\lim_{{x \to -1}} \sqrt{\frac{2x^2 + 2}{8x^2 + 1}}$, we can follow these steps:

Step 1: Substitute the limit value

Substitute $x = -1$ directly into the function:

$\sqrt{\frac{2(-1)^2 + 2}{8(-1)^2 + 1}} = \sqrt{\frac{2(1) + 2}{8(1) + 1}} = \sqrt{\frac{2 + 2}{8 + 1}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$

Final Answer:

$\lim_{{x \to -1}} \sqrt{\frac{2x^2 + 2}{8x^2 + 1}} = \frac{2}{3}$

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How does the behavior of the limit change as $x$ approaches other values, such as 0 or 1?
2. How would you find the limit if the expression inside the square root were more complex?
3. What is the significance of the square root in limit problems?
4. How can the concept of limits be applied to more advanced topics in calculus?
5. How would you evaluate the limit if the function inside the square root was discontinuous at $x = -1$?

Tip: When evaluating limits, always check if direct substitution works first, but be mindful of cases where it might lead to undefined forms.