A calculus problem by A Former Brilliant Member

Calculus Level 3

A continuous and differentiable function f ( x ) f(x) possesses the following limit:

lim ⁡ h → 0 f ( 3 + 7 h ) − f ( 3 + 4 h ) h = 4 \lim_{h \to 0} \dfrac{ f(3+7h) - f(3+4h) }h = 4

What is the value of f ′ ( 3 ) f'(3) ?


The answer is 1.333.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Steven Chase
Jun 8, 2019

Linearize around f ( 3 ) f(3) . In the limit as h h approaches zero:

f ( 3 + 7 h ) − f ( 3 + 4 h ) h = f ( 3 ) + 7 h f ′ ( 3 ) − f ( 3 ) − 4 h f ′ ( 3 ) h = 3 h f ′ ( 3 ) h = 3 f ′ ( 3 ) = 4 f ′ ( 3 ) = 4 3 \frac{f(3 + 7h) - f(3 + 4h)}{h} = \frac{f(3) + 7h \, f'(3) - f(3) - 4h \, f'(3)}{h} \\ = \frac{3 h \, f'(3)}{h} = 3 f'(3) = 4 \\ f'(3) = \frac{4}{3}

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Akshaj Garg
Jun 28, 2019
  1. Apply L'Hôpital's rule.
  2. After putting h = 0 h=0 we can easily get f ′ ( 3 ) = 4 / 3 f'(3)=4/3
0 Helpful 0 Interesting 0 Brilliant 0 Confused

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