Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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)) :}

$$\left({i}\right)\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}+\mathrm{1}}−\frac{\left({ax}+{b}\right)\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}}\right)\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{1}−\left({ax}^{\mathrm{2}} +\left({a}+{b}\right){x}+{b}\right)}{{x}+\mathrm{1}}\right)\:= \\ $$$$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\left(\mathrm{1}−{a}\right){x}^{\mathrm{2}} −\left({a}+{b}\right){x}+\mathrm{1}−{b}}{{x}+\mathrm{1}}\right)\: \\ $$$${because}\:{the}\:{limit}\:{equal}\:{to}\:\mathrm{0}\:,\:{it}\:{meant} \\ $$$$\:\begin{cases}{\mathrm{1}−{a}\:=\:\mathrm{0}\Rightarrow{a}=\mathrm{1}}\\{−\left({a}+{b}\right)=\mathrm{0}\Rightarrow{b}=−\mathrm{1}}\end{cases} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com