Limits of triple differentiation

Calculus Level 4

If $f(x)$ is a thrice differentiable function such that

$\large \lim_{x \rightarrow 0} \frac{f(4x) - 3 f(3x) + 3f(2x) - f(x)}{x^3} = 12,$

find the value of $f'''(0)$ .

The answer is 12.

1 solution

Chew-Seong Cheong
Dec 31, 2016

$\begin{aligned} L & = \lim_{x \to 0} \frac {f(4x)-3f(3x)+3f(2x)-f(x)}{x^3} & \small \color{#3D99F6} \text{A 0/0 cases, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {4f'(4x)-9f'(3x)+6f'(2x)-f'(x)}{3x^2} & \small \color{#3D99F6} \text{A 0/0 cases again, apply L'Hôpital's rule.} \\ & = \lim_{x \to 0} \frac {16f''(4x)-27f''(3x)+12f''(2x)-f''(x)}{6x} & \small \color{#3D99F6} \text{A 0/0 cases again, apply L'Hôpital's rule.} \\ & = \lim_{x \to 0} \frac {64f'''(4x)-81f'''(3x)+24f'''(2x)-f'''(x)}{6} \\ & = f'''(0) = \boxed{12} \end{aligned}$

Isnt it a nice sum?

Md Zuhair - 4 years, 5 months ago

You mean a nice problem. Yes, it is.

Chew-Seong Cheong - 4 years, 5 months ago

Limits of triple differentiation

The answer is 12.

1 solution

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