240927 每日一题

题面

求

$$\underset{x\to 0}{lim}(\frac{2+e^{\frac{1}{x}}}{1+e^{\frac{4}{x}}}+\frac{\sin x}{|x|})$$

解答

$$\begin{array}{rl}& \underset{x\to 0^{+}}{lim}(\frac{2+e^{\frac{1}{x}}}{1+e^{\frac{4}{x}}}+\frac{\sin x}{|x|})\\ & =\underset{x\to 0}{lim}(\frac{2e^{-\frac{4}{x}}+e^{-\frac{3}{x}}}{e^{-\frac{4}{x}}+1}+1)\\ & =\underset{x\to 0}{lim}(\frac{0+0}{0+1}+1)\\ & =1\end{array}$$$$\begin{array}{rl}& \underset{x\to 0^{-}}{lim}(\frac{2+e^{\frac{1}{x}}}{1+e^{\frac{4}{x}}}+\frac{\sin x}{|x|})\\ & =\underset{t\to 0^{+}}{lim}(\frac{2+e^{-\frac{1}{t}}}{1+e^{-\frac{4}{t}}}-\frac{\sin t}{t})\\ & =\underset{t\to 0^{+}}{lim}(\frac{2+0}{1+0}-1)\\ & =1\end{array}$$

综上，

$$\underset{x\to 0}{lim}(\frac{2+e^{\frac{1}{x}}}{1+e^{\frac{4}{x}}}+\frac{\sin x}{|x|})=1$$