Knowing Limits

Calculus Level pending

A function f ( x ) = A x 3 + B x 2 + C x + D x 2 + x − 2 f(x)=\frac{Ax^3+Bx^2+Cx+D}{x^2+x-2} satisfies both lim ⁡ x → 1 f ( x ) = 2 \displaystyle \lim_{x \to 1} f(x)=2 and lim ⁡ x → ∞ f ( x ) = 1 \displaystyle \lim_{x \to \infty} f(x)=1 . What is the value of f ( 4 ) ? f(4)?

5 2 \frac{5}{2} 1 2 \frac{1}{2} 3 2 \frac{3}{2} 7 2 \frac{7}{2}

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Mahindra Jain by Mahindra Jain

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1 solution

Tom Engelsman
Nov 21, 2020

If lim ⁡ x → ∞ f ( x ) = 1 , \lim_{x \rightarrow \infty} f(x) = 1, then we require A = 0 , B = 1. A = 0, B = 1. This now leaves us to figure out C C and D D , and if lim ⁡ x → 1 f ( x ) = 2 \lim_{x \rightarrow 1} f(x) = 2 then we have:

1 + C + D 1 + 1 − 2 = 1 + C + D 0 ⇒ 1 + C + D = 0 \frac{1 + C + D}{1+1-2} = \frac{1+C+D}{0} \Rightarrow 1+C+D = 0

and by L'Hopital's Rule: 2 x + C 2 x + 1 → 2 + C 3 = 2 ⇒ C = 4 ⇒ D = − 5. \frac{2x +C}{2x+1} \rightarrow \frac{2+C}{3} = 2 \Rightarrow C = 4 \Rightarrow D = -5.

Thus, f ( x ) = x 2 + 4 x − 5 x 2 + x − 2 = ( x + 5 ) ( x − 1 ) ( x + 2 ) ( x − 1 ) = x + 5 x + 2 f(x) = \frac{x^2+4x-5}{x^2+x-2} = \frac{(x+5)(x-1)}{(x+2)(x-1)} = \frac{x+5}{x+2} , and f ( 4 ) = 3 2 . f(4) = \boxed{\frac{3}{2}}.

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