Question

$\underset{n\to \infty }{lim}(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}-(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}$ equals

• A. $\frac{A_1-B_1}{2}$
• B. $\frac{A_1+B_3}{2}$
• C. $\frac{B_3-A_1}{2}$
• D. $\frac{A_1-B_3}{2}$

Answer 1

The correct option is A $\frac{A_1-B_1}{2}$

$\underset{n\to \infty }{lim}(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}-(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}$
$=\underset{n\to \infty }{lim}\left[\right(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}-(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}]\times$

$\frac{(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}+(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}}{(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}+(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}}$
$=\underset{n\to \infty }{lim}\frac{(n^4+n^3+A_1{n}^2+A_{2}n+A_3)-(n^4+n^3+B_1{n}^2+B_{2}n+B_3)}{(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}+(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}}$
$=\underset{n\to \infty }{lim}\frac{(A_1-B_1)n^2+(A_2-B_2)n+(A_3-B_3)}{(n^4+n^3+A_1{n}^2+A_{2}n+A_3{)}^{1/2}+(n^4+n^3+B_1{n}^2+B_{2}n+B_3{)}^{1/2}}$
$=\underset{n\to \infty }{lim}\frac{(A_1-B_1)+(A_2-B_2)1/n+(A_3-B_3)/n^2}{(1+1/n+A_1/n^2+A_2/n^3+A_3/n^4{)}^{1/2}+(1+1/n+B_1/n^2+B_2/n^3+B_3/n^4{)}^{1/2}}$, (Divided both Nr and Dr by $n^2$)
$=\frac{A_1-B_1}{1+1}=\frac{A_1-B_1}{2}$