Question

Let f(x) be a twice differentiable function such that f''(0) = 2, then the value of $\lim_{x\to 0} \frac{f(5x) - 3f(3x) + 2f(2x)}{x^2}$

Answer 1

6

Answer 2

The limit is of the indeterminate form $\frac{0}{0}$. Applying L'Hopital's rule twice, we differentiate the numerator and the denominator twice.

The first differentiation of the numerator yields $5f'(5x) - 9f'(3x) + 4f'(2x)$, and the denominator becomes $2x$. This is still $\frac{0}{0}$ at $x=0$.

The second differentiation of the numerator yields $25f''(5x) - 27f''(3x) + 8f''(2x)$, and the denominator becomes $2$. Evaluating the limit as $x \to 0$, we get $\frac{25f''(0) - 27f''(0) + 8f''(0)}{2} = \frac{6f''(0)}{2} = 3f''(0)$. Given $f''(0) = 2$, the limit is $3 \times 2 = 6$.

Alternatively, using the Taylor expansion of $f(x)$ around $x=0$ up to the second order term, $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + O(x^3)$. Substituting $5x$, $3x$, and $2x$ into this expansion and combining the terms in the numerator, we find that the numerator is equal to $3f''(0)x^2 + O(x^3)$. Dividing by $x^2$ and taking the limit as $x \to 0$, we get $3f''(0)$, which evaluates to $3 \times 2 = 6$.