Question

Let $f\left(x\right)$ be a twice differentiable function and $f^{\prime \prime }\left(0\right)=5$, then $\underset{x\to 0}{lim}\frac{3f\left(x\right)-4f\left(3x\right)+f\left(9x\right)}{x^2}$ is equal to:

• A. 30
• B. 120
• C. 40
• D. 5

Answer 1

The correct option is B

120

$=\underset{x\to 0}{lim}\frac{3f\left(x\right)-4f\left(3x\right)+f\left(9x\right)}{x^2}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{3f^{\prime }\left(x\right)-12f^{\prime }\left(3x\right)+9f^{\prime }\left(9x\right)}{2x}\left(\frac{0}{0}form\right)=\underset{x\to 0}{lim}\frac{3f^{\prime \prime }\left(0\right)-36f^{\prime \prime }\left(0\right)+81f^{\prime \prime }\left(9x\right)}{2}=\frac{3f^{\prime \prime }\left(0\right)-36f^{\prime \prime }\left(0\right)+81f^{\prime \prime }\left(0\right)}{2}=24f"\left(0\right)=\left(24\right).\left(5\right)=120$