To find the value of the limit \(\lim_{x\to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^{2}}\), we start by considering a Taylor expansion of \(f(x)\) around \(x = 0\).
The Taylor expansion of a function \(f(x)\) around \(x = 0\) is given by:
\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + \cdots\)
- We have \(f''(0) = k\), so our Taylor expansion simplifies, and particularly the second order term is \(\frac{k}{2}x^2\).
- Substitute \(f(x)\), \(f(2x)\), and \(f(4x)\) using their Taylor expansions:
- \(f(x) = f(0) + f'(0)x + \frac{k}{2}x^2 + \cdots\)
- \(f(2x) = f(0) + f'(0)(2x) + \frac{k}{2}(2x)^2 + \cdots = f(0) + 2f'(0)x + 2kx^2 + \cdots\)
- \(f(4x) = f(0) + f'(0)(4x) + \frac{k}{2}(4x)^2 + \cdots = f(0) + 4f'(0)x + 8kx^2 + \cdots\)
- Substitute these expansions into the expression \(2f(x) - 3f(2x) + f(4x)\):
- \(2f(x) = 2(f(0) + f'(0)x + \frac{k}{2}x^2) = 2f(0) + 2f'(0)x + kx^2 + \cdots\)
- \(-3f(2x) = -3(f(0) + 2f'(0)x + 2kx^2) = -3f(0) - 6f'(0)x - 6kx^2 + \cdots\)
- \(f(4x) = f(0) + 4f'(0)x + 8kx^2 + \cdots\)
- Calculate \(2f(x) - 3f(2x) + f(4x)\):
- \(2f(x) - 3f(2x) + f(4x) = [2f(0) - 3f(0) + f(0)] + [2f'(0)x - 6f'(0)x + 4f'(0)x] + [kx^2 - 6kx^2 + 8kx^2]\)
- The constant and linear terms cancel out, leading to \(3kx^2 + \cdots\) as dominant.
- Divide by \(x^2\):
- \(\frac{2f(x) - 3f(2x) + f(4x)}{x^2} = \frac{3kx^2}{x^2} = 3k\)
Hence, the limit simplifies to:
\(\lim_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2} = 3k\)
Therefore, the correct answer is 3k.