The problem you have shared is asking for the limit of a sum of rational functions as $x$ approaches 2:

$\lim_{{x \to 2}} \left( \frac{x^2 - 8}{x - 2} + \frac{x^2 - 2x}{2x - 4} \right)$

Let's break it down and solve the limit:

Step 1: Factor the expressions

1. For the first term, $\frac{x^2 - 8}{x - 2}$: We can't directly factor $x^2 - 8$ easily, so we'll leave it as is for now.

2. For the second term, $\frac{x^2 - 2x}{2x - 4}$:

   • Factor $2x - 4$ as $2(x - 2)$.
   • The numerator $x^2 - 2x$ can be factored as $x(x - 2)$.

Thus, the second term becomes: $\frac{x(x - 2)}{2(x - 2)}$ Canceling $(x - 2)$, we get: $\frac{x}{2}$

So, the expression now becomes: $\lim_{{x \to 2}} \left( \frac{x^2 - 8}{x - 2} + \frac{x}{2} \right)$

Step 2: Simplify the first term using L'Hopital's Rule

For the first term $\frac{x^2 - 8}{x - 2}$, direct substitution of $x = 2$ leads to an indeterminate form $\frac{0}{0}$. So, we apply L'Hopital's Rule by taking the derivative of the numerator and denominator: $\text{Derivative of } x^2 - 8 = 2x, \quad \text{Derivative of } x - 2 = 1$ So, the first term simplifies to: $\lim_{{x \to 2}} \frac{2x}{1} = 4$

Step 3: Compute the overall limit

Now we compute the limit of the whole expression: $\lim_{{x \to 2}} \left( 4 + \frac{x}{2} \right) = 4 + \frac{2}{2} = 4 + 1 = 5$

Final Answer:

$\lim_{{x \to 2}} \left( \frac{x^2 - 8}{x - 2} + \frac{x^2 - 2x}{2x - 4} \right) = 5$

Would you like more details or have any questions about this solution?

Here are 5 related questions to explore:

1. What is L'Hopital's Rule, and when is it used in limits?
2. How do you handle indeterminate forms in limits without using L'Hopital's Rule?
3. Can all rational functions be simplified using factoring?
4. How can limits help in finding the slope of a curve at a specific point?
5. What are other techniques for solving limits aside from L'Hopital's Rule?

Tip: Always check if you can simplify an expression before applying complex techniques like L'Hopital's Rule.