Evaluate Limit Of Function Using Derivative Values

When faced with a limit problem like $\lim _{x \rightarrow 2} \frac{2 f(2 x)-3 x^2}{4-x^2}$, especially one that provides a table of function and derivative values, our goal is to leverage that information to find the limit. This particular problem requires a careful application of limit properties and, very likely, L'Hôpital's Rule, given its indeterminate form. The table provides us with specific values for $f(x)$ and $f'(x)$ at $x=1$. While $x=1$ might seem distant from the limit point of $x=2$, the structure of the expression often guides us to use these values indirectly. Let's first examine the behavior of the numerator and the denominator as $x$ approaches 2. The denominator, $4-x^2$, clearly approaches $4-2^2 = 0$. Now, let's look at the numerator, $2f(2x) - 3x^2$. As $x \rightarrow 2$, $2x \rightarrow 4$. The problem statement, however, doesn't give us $f(4)$ or $f'(4)$. This suggests we might need to make an assumption or that there's a misunderstanding in how the provided table should be used. Assuming the table intended to provide values relevant to the limit, or that the problem implies a specific function behavior at $x=2$ not explicitly stated, we'd typically check if the numerator also approaches 0. If the numerator approaches a non-zero value while the denominator approaches 0, the limit would be infinite. If both approach 0, we have an indeterminate form $0/0$, which is a strong indicator for L'Hôpital's Rule. Let's proceed with the assumption that the numerator also approaches 0. This would mean $2f(2 \cdot 2) - 3(2)^2 = 0$, so $2f(4) - 12 = 0$, which implies $f(4) = 6$. This value isn't directly in our table, reinforcing the idea that we might be missing context or a key piece of information regarding $f(x)$'s behavior at $x=2$ or $x=4$. However, if we strictly use the provided table and assume the question implicitly requires evaluation at $x=1$ to inform the limit at $x=2$ (which is unusual without further context), we would be stuck. A more standard problem of this type would either provide values at $x=2$ or $x=4$, or imply a function that allows derivation at $x=2$. Let's hypothesize a common scenario for such problems: the table should have included values for $x=2$ or the function $f(x)$ is such that we can deduce $f(2)$ and $f'(2)$ are relevant. Given the limit is as $x \rightarrow 2$, it's most probable that the table is incomplete or the question expects us to realize that L'Hôpital's rule applied to the original expression will eventually require the derivative of the numerator and denominator. The derivative of the denominator $4-x^2$ is $-2x$. As $x \rightarrow 2$, this approaches $-4$. For the numerator $2f(2x) - 3x^2$, its derivative is found using the chain rule: $\frac{d}{dx}(2f(2x)) - \frac{d}{dx}(3x^2) = 2f'(2x) \cdot 2 - 6x = 4f'(2x) - 6x$. Now, applying L'Hôpital's Rule, the limit becomes $\lim _{x \rightarrow 2} \frac{4f'(2x) - 6x}{-2x}$. This form requires $f'(4)$ to be known. If the table is indeed for $x=1$ only, and no other values are given or implied, the problem as stated is unsolvable without additional information or clarification. It's highly likely that the problem statement or the provided table is flawed. A standard solvable version would either give $f(2)$ and $f'(2)$ or $f(4)$ and $f'(4)$, or perhaps a function definition. Let's assume, for the sake of demonstrating the method, that the table should have contained values for $x=2$. If we had $f(2)$ and $f'(2)$, and assuming the indeterminate form $0/0$ holds, we would evaluate $\frac{4f'(2 \cdot 2) - 6 \cdot 2}{-2 \cdot 2} = \frac{4f'(4) - 12}{-4}$. This still requires $f'(4)$. If, however, the limit was $x \rightarrow 1$, and the table provided values for $x=1$, then the expression would be $\lim _{x \rightarrow 1} \frac{2 f(2 x)-3 x^2}{4-x^2}$. The denominator approaches $4-1^2 = 3$. The numerator approaches $2f(2) - 3(1)^2$. We don't have $f(2)$. This confirms the original problem statement's table seems mismatched with the limit's target value. Let's re-evaluate the possibility that the table values at $x=1$ are meant to be used in a way that's not immediately obvious, perhaps through some property not commonly recalled. However, in standard calculus, limits involving derivatives at a point 'a' typically require function/derivative values at 'a' or points that can be directly substituted into the differentiated expression. The structure $\frac{2 f(2 x)-3 x^2}{4-x^2}$ as $x \rightarrow 2$ points to needing information about $f$ and its derivative at $x=4$ (from $2x$) and at $x=2$ (from $x^2$ and $-x^2$). If the question implies $f(x)$ is a simple polynomial, we might be able to infer values, but without that, we rely on the provided data. The most plausible interpretation is that the table is incomplete and should have included values for $x=2$ or $x=4$. Let's assume the question meant for the limit to be as $x \rightarrow 1$, and the table provided values at $x=1$. In that hypothetical scenario: $\lim _{x \rightarrow 1} \frac{2 f(2 x)-3 x^2}{4-x^2}$. Denominator $\rightarrow 4-1^2 = 3$. Numerator $\rightarrow 2f(2) - 3(1)^2$. We don't have $f(2)$. This hypothetical also fails. The problem as stated, with the table provided, appears to have a disconnect.

Let's reconsider the core problem: $\lim _{x \rightarrow 2} \frac{2 f(2 x)-3 x^2}{4-x^2}$. The denominator $4-x^2 \rightarrow 0$ as $x \rightarrow 2$. For the limit to be a finite value (as implied by