Question

If $f''(0) = k$, then $\displaystyle\lim_{x\to 0} \dfrac{2f(x) - 3f(2x) + f(4x)}{x^{2}}$ equals

• A. $k$
• B. $2k$
• C. $3k$
• D. $4k$

Hint:

For limits of the form $[2f(x)-3f(2x)+f(4x)]/x^2$, expand via Taylor series — the $f(0)$ and $f'(0)$ terms cancel, and only the $f''(0)$ term contributes.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Expand each term using the Taylor series about $x=0$ up to the $x^2$ term.
Step 2: Detailed Explanation:
$f(x) \approx f(0)+xf'(0)+\tfrac{x^2}{2}f''(0)$.
Coefficient of $x^2/2$ in $2f(x)-3f(2x)+f(4x)$: $2(1)-3(4)+16 = 2-12+16=6$.
Numerator $\approx 6\cdot\dfrac{x^2}{2}f''(0) = 3kx^2$. Divide by $x^2$: limit $= 3k$.
Step 3: Final Answer:
The limit equals $3k$.