Question Number 199467 by Calculusboy last updated on 04/Nov/23
Answered by cortano12 last updated on 04/Nov/23
![4 = lim_(x→∞) (((a−2)x^3 +(3+c)x^2 +(b−3)x+2+d)/(x^2 [(√(1+(a/x)+(3/x^2 )+(b/x^3 )+(2/x^4 ))) +(√(1+(2/x)−(c/x^2 )+(3/x^3 )−(d/x^4 ))) ])) { ((a−2=0⇒a=2)),((((3+c)/2)=4⇒c=5 )),((b−3=0⇒b=3)),((2+d=0⇒d=−2)) :}](https://www.tinkutara.com/question/Q199472.png)
$$\:\mathrm{4}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{a}−\mathrm{2}\right)\mathrm{x}^{\mathrm{3}} +\left(\mathrm{3}+\mathrm{c}\right)\mathrm{x}^{\mathrm{2}} +\left(\mathrm{b}−\mathrm{3}\right)\mathrm{x}+\mathrm{2}+\mathrm{d}}{\mathrm{x}^{\mathrm{2}} \:\left[\sqrt{\mathrm{1}+\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{b}}{\mathrm{x}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{x}^{\mathrm{4}} }}\:+\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{x}}−\frac{\mathrm{c}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{3}} }−\frac{\mathrm{d}}{\mathrm{x}^{\mathrm{4}} }}\:\right]} \\ $$$$\:\begin{cases}{\mathrm{a}−\mathrm{2}=\mathrm{0}\Rightarrow\mathrm{a}=\mathrm{2}}\\{\frac{\mathrm{3}+\mathrm{c}}{\mathrm{2}}=\mathrm{4}\Rightarrow\mathrm{c}=\mathrm{5}\:}\\{\mathrm{b}−\mathrm{3}=\mathrm{0}\Rightarrow\mathrm{b}=\mathrm{3}}\\{\mathrm{2}+\mathrm{d}=\mathrm{0}\Rightarrow\mathrm{d}=−\mathrm{2}}\end{cases} \\ $$