Question

$\underset{x\to \infty }{lim}[(x^2+\frac{1}{x})e^{1/x}-x-x^2]$ is equal to

• A. $\frac{1}{4}$
• B. $2$
• C. $\frac{1}{8}$
• D. $\frac{1}{2}$

Answer 1

The correct option is D $\frac{1}{2}$

$\underset{x\to \infty }{lim}[(x^2+\frac{1}{x})e^{\frac{1}{x}}-x-x^2]$
Put $x=\frac{1}{t}$
$=\underset{t\to 0}{lim}[(\frac{1}{t^2}+t)e^t-\frac{1}{t}-\frac{1}{t^2}]$
$=\underset{t\to 0}{lim}\frac{t^3{e}^t+e^t-t-1}{t^2}$
$=\underset{t\to 0}{lim}\frac{t^3(1+t+\frac{t^2}{2!}+\cdots )+(1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots )-t-1}{t^2}$
$=\underset{t\to 0}{lim}\frac{t^3(1+t+\frac{t^2}{2!}+\cdots )+(\frac{t^2}{2!}+\frac{t^3}{3!}+\cdots )}{t^2}$
$=\underset{t\to 0}{lim}[t(1+t+\frac{t^2}{2!}+\cdots )+\frac{1}{2}+\frac{t}{3!}+\cdots ]$
$=\frac{1}{2}$