Question

The value of $\underset{x\to 0}{lim}\frac{(1+x{)}^{1/x}-e+\frac{1}{2}ex}{x^2}$ is

• A. $\frac{11}{24}e$
• B. $-\frac{11}{24}e$
• C. $\frac{e}{24}$
• D. none of these

Answer 1

The correct option is A $\frac{11}{24}e$

Let $y=(1+x{)}^{1/x}$
$\Rightarrow logy=\frac{1}{x}log(1+x)=\frac{1}{x}[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.....]$
$=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+....$
$y=e^{(1-\frac{x}{2}+\frac{x^2}{3}-....)}=e.e^{(-\frac{x}{2}+\frac{x^2}{3}-....)}$
$=e[1+(-\frac{x}{2}+\frac{x^2}{3}-...)+\frac{1}{2!}{(-\frac{x}{2}+\frac{x^2}{3}-)}^2+...]$
$y-e+\frac{1}{2}ex=e{x}^2[\frac{1}{3}+0(x)+\frac{1}{2}{(-\frac{1}{2}+0(x\left)\right)}^2+...]$
($0\left(x\right)$ in terms containing x)
$\underset{x\to 0}{lim}\frac{y-e+\frac{1}{2}ex}{x^2}=e[\frac{1}{3}+\frac{1}{8}]=\frac{11}{24}e$