A calculus problem - 2

Calculus Level 4

If f ( x ) f(x) is a polynomial satisfying lim ⁡ x → ∞ x f ( x ) x 5 + 1 = 1 \displaystyle \lim _{ x\to \infty }{ \dfrac { xf(x) }{ { x }^{ 5 }+1 } } =1 , f ( 1 ) = 1 f(1)=1 , f ( 2 ) = 2 f(2)=2 , f ( 3 ) = 3 f(3)=3 and f ( 4 ) = 4 f(4)=4 then what is lim ⁡ x → 3 x − f ( x ) x − 3 \displaystyle \lim _{ x\to3 }{ \frac { x-f(x) }{ x-3 } } ?


The answer is 2.

Swagat Panda by Swagat Panda , 22, India

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1 solution

Sabhrant Sachan
Jul 9, 2016

The Limit tells us that the function f ( x ) must be a polynomial of degree 4 with leading coefficient 1 Since f ( 1 ) = 1 , f ( 2 ) = 2 , f ( 3 ) = 3 and f ( 4 ) = 4 \text{The Limit tells us that the function } f(x) \text{ must be a polynomial of degree } 4 \text{ with leading coefficient } 1 \\ \text{Since } f(1)=1 , f(2) =2 , f(3) = 3 \text{ and } f(4) = 4

f ( x ) = ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) + x lim ⁡ x → 3 x − f ( x ) x − 3 lim ⁡ x → 3 − 1 ⋅ ( x − 1 ) ( x − 2 ) ( x − 4 ) = 2 f(x) = (x-1)(x-2)(x-3)(x-4)+x \\ \displaystyle \lim_{x \to 3} \dfrac{x-f(x)}{x-3} \\ \displaystyle\lim_{x \to 3} -1\cdot (x-1)(x-2)(x-4) \\ = \boxed{2}

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