Question

${lim}_{x\to 0}\frac{(1+x{)}^{\frac{1}{x}}+e(x-1)}{si{n}^{-1}x}=$

• A. $-\frac{1}{3}$
• B. $\frac{e}{2}$
• C. $\frac{1}{2e}$
• D. $\frac{e^2}{8}$

Answer 1

The correct option is B $\frac{e}{2}$

$L={lim}_{x\to 0}\frac{(1+x{)}^{\frac{1}{x}}+e(x-1)}{x}\times \frac{x}{si{n}^{-1}x}\Rightarrow L={lim}_{x\to 0}(\frac{e^{\frac{ln(1+x)}{x}}-e}{x}+e)\times \left({lim}_{x\to 0}\frac{x}{si{n}^{-1}}x\right)=e\times {lim}_{x\to 0}(\frac{e^{-\frac{x}{2}+\frac{x^2}{3}\dots }-1}{x}+1)\times 1=e\times (-\frac{1}{2}+1)=\frac{e}{2}$