Question

Evaluate $\underset{x\to 0}{\lim }\frac{(1+x{)}^{1/x}-e+\frac{1}{2}ex}{x^2}$.

• A. $\frac{11e}{24}$
• B. $e$
• C. $\frac{7e}{19}$
• D. $1$

Answer 1

Answer: (A)

$(1+x{)}^{1/x}=e^{\frac{1}{x}\log (1+x)}=e^{\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\dots \right)}$
$=e^{1-\frac{x}{2}+\frac{x^2}{3}-\dots .}=e\cdot e^{-\frac{x}{2}+\frac{x^2}{3}-\dots }$
$=e\left[1+\left(-\frac{x}{2}+\frac{x^2}{3}-\dots .\right)+\frac{1}{2!}{\left(-\frac{x}{2}+\frac{x^2}{3}\dots .\right)}^2+\dots .\right]$
$=e\left[1-\frac{x}{2}+\frac{11}{24}{x}^2-\dots ..\right]$
Hence, $\underset{x\to 0}{\lim }\frac{(1+x{)}^{1/x}-e+\frac{1}{2}ex}{x^2}=\frac{11e}{24}$