The image contains the following limit problem:

$\lim_{{x \to 2}} \frac{2x^2 - 16}{x - 2}$

Solution:

1. Factor the numerator: We first need to factor $2x^2 - 16$. Notice that:

   $2x^2 - 16 = 2(x^2 - 8)$

   However, this doesn't seem to simplify in a helpful way for this particular limit. Another strategy is using L'Hopital's Rule, since directly substituting $x = 2$ results in the indeterminate form $\frac{0}{0}$.

2. Apply L'Hopital's Rule: L'Hopital's Rule states that if we get an indeterminate form $\frac{0}{0}$, we can take the derivative of the numerator and the denominator.

   • Derivative of the numerator $2x^2 - 16$: $\frac{d}{dx}(2x^2 - 16) = 4x$
   • Derivative of the denominator $x - 2$: $\frac{d}{dx}(x - 2) = 1$

3. Evaluate the limit: Now, using L'Hopital's Rule, we take the limit of the derivatives:

   $\lim_{{x \to 2}} \frac{4x}{1} = 4 \times 2 = 8$

Thus, the solution to the limit is:

$\boxed{8}$

Would you like further details on this process or have any questions?

Here are 5 related questions that you can explore:

1. How does factoring help in solving limits?
2. What are the conditions for using L'Hopital's Rule?
3. How do you factor more complex quadratic expressions?
4. Can limits be evaluated using methods other than L'Hopital’s Rule?
5. What happens if the limit is not indeterminate (not $\frac{0}{0}$)?

Tip: Always check if direct substitution works before applying L'Hopital’s Rule. If it doesn't, check if the expression simplifies or use advanced techniques like factoring or series expansion.