Question

$f\left(x\right)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5.$ If $f\left(x\right)$ satisfies $\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x^2}]}^{1/x}=e^3$, then which of the following is correct?

• A. $a_0=0,a_3=3$
• B. $a_0=0,a_3=4$
• C. $a_1=1,a_3=3$
• D. $a_0=1,a_4=3$

Answer 1

The correct option is A $a_0=0,a_3=3$

$f\left(x\right)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5$
$\underset{x\to 0}{lim}{[1+\frac{f\left(x\right)}{x^2}]}^{1/x}=e^3$
$\Rightarrow e^{\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}}=e^3\left[1^{\infty }\text{form}\right]$
$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=3$
$\underset{x\to 0}{lim}\frac{a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+a_5{x}^5}{x^3}=3$
$\underset{x\to 0}{lim}\frac{a_0}{x^3}+\frac{a_1}{x^2}+\frac{a_2}{x}+a_3+a_{4}x+a_5{x}^2=3$

Therefore, for limit to exist at $x=0$,$a_0=a_1=a_2=0$ and $a_3=3$ whereas $a_4$ and $a_5$ can take any real values.