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Question Number 94214 by abony1303 last updated on 17/May/20

Find f(0) when a polynomial f(x) satisfies  lim_(x→1)  ((f(x))/(x^2 −1))=2  lim_(x→−1)   ((f(x))/(x^2 −1))=2    lim_(x→+∞)   ((f(x))/x^4 ) =1        pls Help!

$${Find}\:{f}\left(\mathrm{0}\right)\:{when}\:{a}\:{polynomial}\:{f}\left({x}\right)\:{satisfies} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{2}\:\: \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\:\frac{{f}\left({x}\right)}{{x}^{\mathrm{4}} }\:=\mathrm{1}\:\:\:\:\:\:\:\:{pls}\:{Help}! \\ $$

Commented by john santu last updated on 17/May/20

lim_(x→+∞)  ((f(x))/x^4 ) = ??

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{4}} }\:=\:?? \\ $$

Commented by john santu last updated on 17/May/20

f(x) = (x^2 −1)(ax^2 +b)

$$\mathrm{f}\left({x}\right)\:=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({ax}^{\mathrm{2}} +{b}\right)\: \\ $$$$ \\ $$

Commented by abony1303 last updated on 17/May/20

changed

$${changed} \\ $$

Commented by i jagooll last updated on 17/May/20

f(x)=(x^2 −1)(ax^2 +bx+c)  (1) lim_(x→1)  (((x^2 −1)(ax^2 +bx+c))/(x^2 −1)) = 2  ⇒a+b+c = 2  (2) lim_(x→−1)  (((x^2 −1)(ax^2 +bx+c))/(x^2 −1)) = 2  ⇒ a−b+c = 2  (3) lim_(x→+∞) ((f(x))/x^4 ) = 1 ⇒ a = 1   eq (1)+(2) ⇒ 2a+2c = 4 ⇒ c = 1 and   b = 0 . then f(x) = (x^2 −1)(x^2 +1)  f(0) = −1

$${f}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({ax}^{\mathrm{2}} +{bx}+{c}\right) \\ $$$$\left(\mathrm{1}\right)\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\Rightarrow{a}+{b}+{c}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\underset{{x}\rightarrow−\mathrm{1}} {\mathrm{lim}}\:\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({ax}^{\mathrm{2}} +{bx}+{c}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}\:=\:\mathrm{2} \\ $$$$\Rightarrow\:{a}−{b}+{c}\:=\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{f}\left({x}\right)}{{x}^{\mathrm{4}} }\:=\:\mathrm{1}\:\Rightarrow\:{a}\:=\:\mathrm{1}\: \\ $$$${eq}\:\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\Rightarrow\:\mathrm{2}{a}+\mathrm{2}{c}\:=\:\mathrm{4}\:\Rightarrow\:{c}\:=\:\mathrm{1}\:{and}\: \\ $$$${b}\:=\:\mathrm{0}\:.\:{then}\:{f}\left({x}\right)\:=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$${f}\left(\mathrm{0}\right)\:=\:−\mathrm{1} \\ $$$$ \\ $$

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