Question

$\underset{x\to 0}{\text{lim}}\frac{{(1+x)}^{\frac{1}{x}}-e(1-\frac{x}{2})}{(1-\cos x)}=$

• A. $\frac{1}{2}e$
• B. $\frac{1}{4}e$
• C. $\frac{11}{12}e$
• D. $\frac{1}{12}e$

Answer 1

The correct option is C $\frac{11}{12}e$

$(1+x{)}^{\frac{1}{x}}=e^{\frac{1}{x}\log (1+x)}$
$=e^{(x-\frac{x^2}{2}+\frac{x^3}{3}\cdots )\frac{1}{x}}$
$=e^{(1-\frac{x}{2}+\frac{x^2}{3}\cdots )}$
$=e\cdot e^{-m}(m=\frac{x}{2}-\frac{x^2}{3}\cdots )$

$=e(1-m+\frac{m^2}{2!}-\frac{m^3}{3!}+\cdots )$

$=e(1-(\frac{x}{2}-\frac{x^2}{3}+\cdots )+\frac{{(\frac{x}{2}-\frac{x^3}{3}+\cdots )}^2}{2!}-\cdots )$

$=e(1-\frac{x}{2}+\frac{11}{24}{x}^2+...)$

Now, the given limit can be written as:
$\underset{x\to 0}{\text{lim}}\frac{e(1-\frac{x}{2}+\frac{11}{24}{x}^2+...)-e(1-\frac{x}{2})}{(1-\cos x)}$

= $\underset{x\to 0}{\text{lim}}\frac{\frac{11}{24}e{x}^2}{2{\sin }^2\left(\frac{x}{2}\right)}=\frac{11e}{12}{\left(\frac{\frac{x}{2}}{\sin \frac{x}{2}}\right)}^2=\frac{11}{12}e$