The limit you're trying to evaluate is:

$\lim_{x \to 5} \frac{50 - 2x^2}{x - 5}$

To solve this, follow these steps:

Step 1: Direct Substitution

Substitute $x = 5$ directly into the expression:

$\frac{50 - 2(5)^2}{5 - 5} = \frac{50 - 50}{0} = \frac{0}{0}$

Since this results in an indeterminate form $\frac{0}{0}$, we need to apply further techniques like factoring, simplifying, or using L'Hôpital's Rule.

Step 2: Factorization or Simplification

Let's first simplify the numerator $50 - 2x^2$. We observe that:

$50 - 2x^2 = 2(25 - x^2)$

We can factor $25 - x^2$ as a difference of squares:

$25 - x^2 = (5 - x)(5 + x)$

So, the expression becomes:

$\frac{2(5 - x)(5 + x)}{x - 5}$

Notice that $5 - x$ can be rewritten as $-(x - 5)$, so:

$\frac{2(-(x - 5))(5 + x)}{x - 5} = \frac{-2(x - 5)(5 + x)}{x - 5}$

The $x - 5$ terms cancel out (since $x \neq 5$):

$\lim_{x \to 5} -2(5 + x)$

Step 3: Evaluate the Limit

Substitute $x = 5$ into the simplified expression:

$-2(5 + 5) = -2(10) = -20$

Final Answer:

$\lim_{x \to 5} \frac{50 - 2x^2}{x - 5} = -20$

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

Here are some related questions to expand your understanding:

1. What are the conditions under which L'Hôpital's Rule can be applied to solve limits?
2. How do you factor the difference of squares in general?
3. What is the importance of checking for indeterminate forms when evaluating limits?
4. Can limits always be solved by direct substitution?
5. How does the behavior of a function near a point affect the limit at that point?

Tip: When you encounter an indeterminate form like $\frac{0}{0}$ in limits, consider factoring, simplifying, or applying L'Hôpital's Rule to resolve it.