Question

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

• A. $\frac{11}{24}$
• B. $\frac{11e}{24}$
• C. $\frac{24e}{11}$
• D. $\frac{12e}{11}$

Answer 1

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

$F\left(x\right)=(1+x{)}^{\frac{1}{x}}-e+\frac{ex}{2}$

$=exp\left(\frac{\log (1+x)}{x}\right)-exp(1+\log (1-\frac{x}{2}))$

$=exp\{1+\log (1-\frac{x}{2})\}\{exp(\frac{\log (1+x)}{x}-1-\log (1-\frac{x}{2}))-1\}$

$=e^u(e^v-1)$

$u=1+\log (1-\frac{x}{2})$

$v=\frac{\log (1+x)}{x}-1-\log (1-\frac{x}{2})$

when $x\to 0,u\to 1,v\to 0$

$L={lim}_{x\to 0}\frac{F\left(x\right)}{x^2}$

$={lim}_{x\to 0}\frac{e^u(e^v-1)}{v}\cdot \frac{v}{x^2}$

$=e\cdot {lim}_{x\to 0}\frac{1}{x^2}\{\frac{\log (1+x)}{x}-1-\log (1-\frac{x}{2})\}$

$=e\cdot {lim}_{x\to 0}\frac{1}{x^2}\{(1-\frac{x}{2}+\frac{x^2}{3-......})-1+(\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{24}+...)\}$

$=e(\frac{1}{3}+\frac{1}{8})$

$=\frac{11e}{24}$