Question

Let $f:R \rightarrow R$ be a function such that $f(2)=4$ and $f'(2)=1$. Then, the value of $\lim_{x\to 2}\frac{x^2f(2) - 4f(x)}{x-2}$ is equal to :

• A. 4
• B. 8
• C. 12
• D. 16

Hint:

Whenever a limit involves $f(x)$ and $f(a)$ divided by $(x-a)$, try to rewrite the expression to directly use the definition of derivative: \[ f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \] This method is often faster and more reliable than L'Hôpital’s Rule.

Answer 1

The Correct Option is C

We are given: \[ f(2)=4 \quad \text{and} \quad f'(2)=1 \] We need to evaluate: \[ L = \lim_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x-2} \] Step 1: Check the form of the limit
Substitute $x=2$: \[ \text{Numerator} = 2^2 f(2) - 4f(2) = 4f(2) - 4f(2) = 0 \] \[ \text{Denominator} = 2 - 2 = 0 \] So, the limit is of the indeterminate form $\frac{0}{0}$. Step 2: Rearrange the expression
\[ \frac{x^2 f(2) - 4f(x)}{x-2} = \frac{(x^2 - 4)f(2) - 4[f(x)-f(2)]}{x-2} \] Factor where possible: \[ = \frac{(x-2)(x+2)f(2)}{x-2} - 4 \cdot \frac{f(x)-f(2)}{x-2} \] Cancel $(x-2)$: \[ = (x+2)f(2) - 4 \cdot \frac{f(x)-f(2)}{x-2} \] Step 3: Take the limit as $x \to 2$
Using $f(2)=4$ and the definition of derivative: \[ \lim_{x\to 2}\frac{f(x)-f(2)}{x-2} = f'(2) = 1 \] So, \[ L = (2+2)\cdot 4 - 4\cdot 1 \] \[ L = 16 - 4 = 12 \] \[ \boxed{\text{Correct Answer: (C) }12} \]