Question

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

• A. 0
• B. 120
• C. -120
• D. does not exist

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\frac{3f(x)-4f(3x)+f(9x)}{x^2}(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{3f^{\prime }(x)-12f^{\prime }(3x)+9f^{\prime }(9x)}{2x}(\frac{0}{0}$ form $)$
$=\underset{x\to 0}{\lim }\frac{3f^{\prime \prime }(x)-36f^{\prime \prime }(3x)+81f^{\prime \prime }(9x)}{2}$
$=\frac{3f^{\prime \prime }(0)-36f^{\prime \prime }(0)+81f^{\prime \prime }(0)}{2}$
$=24f^{\prime }(0)=24.5=120$