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Find-the-constants-a-and-b-from-the-condition-i-lim-x-x-2-1-x-1-ax-b-0-ii-lim-x-x-2-x-1-ax-b-0-

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Question Number 122725 by liberty last updated on 19/Nov/20
Find the constants a and b from the condition : (i) lim_(x→∞) (((x^2 +1)/(x+1))−ax−b)=0 (ii) lim_(x→∞) ((√(x^2 −x+1))−ax−b)=0
$$\:{Find}\:{the}\:{constants}\:{a}\:{and}\:{b}\:{from}\:{the} \\ $$$${condition}\::\: \\ $$$$\left({i}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}+\mathrm{1}}−{ax}−{b}\right)=\mathrm{0} \\ $$$$\left({ii}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−{ax}−{b}\right)=\mathrm{0} \\ $$
Commented by Dwaipayan Shikari last updated on 19/Nov/20
lim_(x→∞) (√(x^2 −x+1))−ax−b lim_(x→∞) x((√(1−(1/x)+(1/x^2 )))−a−(b/x)) =x(1−(1/(2x))+(1/(2x^2 ))−a−(b/x))=0 If 1−a=0⇒a=1 and then it becomes −(1/2)+(1/(2x))−b=0⇒ x→∞ b=−(1/2) so { ((a=1)),((b=−(1/2))) :}
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}−{ax}−{b} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}\left(\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−{a}−\frac{{b}}{{x}}\right) \\ $$$$={x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }−{a}−\frac{{b}}{{x}}\right)=\mathrm{0} \\ $$$${If}\:\mathrm{1}−{a}=\mathrm{0}\Rightarrow{a}=\mathrm{1} \\ $$$${and}\:{then}\:{it}\:{becomes}\:−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{x}}−{b}=\mathrm{0}\Rightarrow\:\:\: \\ $$$${x}\rightarrow\infty\:\:\:{b}=−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:{so}\:\begin{cases}{{a}=\mathrm{1}}\\{{b}=−\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$
Commented by bemath last updated on 19/Nov/20
(i) lim_(x→∞) (((x^2 +1)/(x+1))−(((ax+b)(x+1))/(x+1))) = lim_(x→∞) (((x^2 +1−(ax^2 +(a+b)x+b))/(x+1))) = lim_(x→∞) ((((1−a)x^2 −(a+b)x+1−b)/(x+1))) because the limit equal to 0 , it meant { ((1−a = 0⇒a=1)),((−(a+b)=0⇒b=−1)) :}
$$\left({i}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}+\mathrm{1}}−\frac{\left({ax}+{b}\right)\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}\right)\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}−\left({ax}^{\mathrm{2}} +\left({a}+{b}\right){x}+{b}\right)}{{x}+\mathrm{1}}\right)\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{1}−{a}\right){x}^{\mathrm{2}} −\left({a}+{b}\right){x}+\mathrm{1}−{b}}{{x}+\mathrm{1}}\right)\: \\ $$$${because}\:{the}\:{limit}\:{equal}\:{to}\:\mathrm{0}\:,\:{it}\:{meant} \\ $$$$\:\begin{cases}{\mathrm{1}−{a}\:=\:\mathrm{0}\Rightarrow{a}=\mathrm{1}}\\{−\left({a}+{b}\right)=\mathrm{0}\Rightarrow{b}=−\mathrm{1}}\end{cases} \\ $$

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