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The problem is asking for the value of \( f'(5) \), where the function \( f(x) \) satisfies the given limit conditions:

1. \( \lim_{x \to \infty} \frac{f(x) - 2x^2}{x^2 - 1} = 2 \)
2. \( \lim_{x \to 1} \frac{f(x) - 2x^2}{x^2 - 1} = 2 \)

### Step 1: Analyze the first limit (as \( x \to \infty \))
We are given:
\[
\lim_{x \to \infty} \frac{f(x) - 2x^2}{x^2 - 1} = 2
\]

This suggests that \( f(x) \) behaves like \( 2x^2 \) as \( x \to \infty \), but we need to account for any potential lower-degree terms. Let's assume \( f(x) \) takes the form:
\[
f(x) = 2x^2 + ax + b
\]
We now substitute this into the limit expression:
\[
\frac{(2x^2 + ax + b) - 2x^2}{x^2 - 1} = \frac{ax + b}{x^2 - 1}
\]
As \( x \to \infty \), this simplifies to:
\[
\frac{ax + b}{x^2} \approx \frac{a}{x} + \frac{b}{x^2}
\]
Since the limit is 2, we must have \( a = 0 \) (as the leading term would otherwise blow up), meaning:
\[
f(x) = 2x^2 + b
\]
Thus, the first limit tells us that \( f(x) = 2x^2 + b \).

### Step 2: Analyze the second limit (as \( x \to 1 \))
We are given:
\[
\lim_{x \to 1} \frac{f(x) - 2x^2}{x^2 - 1} = 2
\]
Substitute \( f(x) = 2x^2 + b \) into the expression:
\[
\frac{(2x^2 + b) - 2x^2}{x^2 - 1} = \frac{b}{x^2 - 1}
\]
As \( x \to 1 \), the denominator becomes 0, so for the limit to exist, we use the fact that \( x^2 - 1 = (x-1)(x+1) \). Thus:
\[
\lim_{x \to 1} \frac{b}{(x-1)(x+1)} = 2
\]
Evaluating this limit as \( x \to 1 \) gives:
\[
\frac{b}{2} = 2 \quad \Rightarrow \quad b = 4
\]

### Step 3: Derive \( f'(x) \) and find \( f'(5) \)
From the above, we have \( f(x) = 2x^2 + 4 \). Taking the derivative:
\[
f'(x) = 4x
\]
Thus:
\[
f'(5) = 4 \times 5 = 20
\]

The value of \( f'(5) \) is \( \boxed{20} \).

---

Would you like more details, or do you have any further questions?

Here are five related questions for further understanding:
1. How do limits help determine the form of a function?
2. What happens if the limits were different at infinity and near a specific point?
3. Why is it necessary to account for lower-order terms in function approximations?
4. How does the derivative change with different functional forms?
5. Can you explain how to apply L'Hopital's Rule to limits involving indeterminate forms?

**Tip**: When working with limits, always look for patterns or terms that dominate the behavior of the function as \( x \to \infty \) or near specific points.

다항함수 f(x)가 다음 조건을 만족시킬 때, f'(5)의 값은?

(1) \( \lim_{x \to \infty} \frac{f(x) - 2x^2}{x^2 - 1} = 2 \)
(2) \( \lim_{x \to 1} \frac{f(x) - 2x^2}{x^2 - 1} = 2 \)

Solve for f'(5) of a polynomial function with given limit conditions. Learn how limits reveal function behavior and how differentiation applies to find specific values. This solution involves polynomial approximation, limit evaluations, and derivative rules, suitable for higher-education students.

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