题目

$$
\text{求极限 } \lim_{x\to0}\frac{(1+x)^{\frac{1}{x}} - (1 + 2x)^{\frac{1}{2x}}}{\sin x}
$$

解答

$$
\begin{aligned}
\lim_{x\to0}\frac{(1+x)^{\frac{1}{x}} - (1 + 2x)^{\frac{1}{2x}}}{\sin x}
&=
\lim_{x\to0}\frac{e^{\frac{\ln(x + 1)}{x}} - e^{\frac{\ln(2x + 1)}{2x}}}{x}
\\
&=
\lim_{x\to0} e^{\frac{\ln(2x + 1)}{2x}} \cdot
\frac{e^{\frac{2\ln(x + 1) - \ln(2x + 1)}{2x}} - 1}{x}
\\
&=
e\cdot
\lim_{x\to0}
\frac{2\ln(x + 1) - \ln(2x + 1)}{2x^2}
\\
&=
e\cdot
\lim_{x\to0}
\frac{2x -x^2 - 2x + 2x^2 + o(x^2)}{2x^2}
\\
&=
\frac{e}{2}
\end{aligned}
$$