Question

If $\underset{x\to \infty }{lim}{(1+\frac{a}{x}+\frac{b}{x^2})}^{2x}$= $e^2$, then the values of $a$ and $b$, are

• A. $a\in R,b\in R$
• B. $a=1,b\in R$
• C. $a\in R,b=2$
• D. $a=1,b=2$

Answer 1

Answer: (B)

$a=1,b\in R$

$\underset{x\to \infty }{lim}{(1+\frac{a}{x}+\frac{b}{x^2})}^{2x}=\underset{x\to \infty }{lim}{(1+\frac{a}{x}+\frac{b}{x^2})}^{2x\left(\frac{\frac{a}{x}+\frac{b}{x^2}}{\frac{a}{x}+\frac{b}{x^2}}\right)}=\underset{x\to \infty }{lim}{e}^{2x(\frac{a}{x}+\frac{b}{x^2})}=\underset{x\to \infty }{lim}{e}^{\left(2(a+\frac{b}{x})\right)}=e^{2a}\Rightarrow a=1,b\in R$